Answer:
V = 411.43 V
Explanation:
The two forces as a result of each of the 2 charges are;
F1 = kq1•q/r
F2 = kq2.q/r
Where r = r/2 since we are dealing with potential difference at a point midway between the charges.
q1 = 5 nC = 5 × 10^(-9) C
q2 = 3 nC = 3 × 10^(-9) C
k = 9 × 10^(9) N.m²/C²
r = 35 cm = 0.35m
r/2 = 0.35/2
Thus;
F1 = (9 × 10^(9) × 5 × 10^(-9) × q)/(0.35/2)²
F1 = 1469.39q
F2 = (9 × 10^(9) × 3 × 10^(-9) × q)/(0.35/2)²
F2 = 881.63q
Net force acting midway is;
F_net = F1 + F2
F_net = 1469.39q + 881.63q
F_net = 2351.02q
Now, we know that formula for electric potential is;
V = kq/r
Thus ;
V = Fr/q derived from the earlier equation for force we used.
Where F is F_net.
V = 2351.02q × r/q
V = 2351.02r
Recall that we are dealing with midpoint and r = r/2
Thus;
V = 2351.02 × 0.35/2
V = 411.43 V
A snail is traveling along a straight path. The snail's velocity can be modeled by v(t) = 1.4 In (1 +r²) inches per minute for 0 ≤ 1 ≤ 15 minutes. (a) Find the acceleration of the snail at time t = 5 minutes. (b) What is the displacement of the snail over the interval 0 ≤ 1 ≤ 15 minutes?
The snail's acceleration at t = 5 minutes is approximately 0.079 inches per minute squared. Over the interval 0 ≤ t ≤ 15 minutes, the snail's displacement is approximately 15.405 inches.
To find the acceleration at t = 5 minutes, we need to differentiate the velocity function with respect to time. The given velocity function is v(t) = 1.4 ln(1 + r²), where ln denotes the natural logarithm. Let's differentiate v(t) with respect to t to find the acceleration function a(t):
a(t) = d/dt (1.4 ln(1 + r²))
To differentiate ln(1 + r²), we use the chain rule:
a(t) = 1.4 * d/dt (ln(1 + r²))
The derivative of ln(1 + r²) with respect to r² is 1 / (1 + r²), so we can rewrite the acceleration function as:
a(t) = 1.4 * (1 / (1 + r²)) * d/dt (1 + r²)
The derivative of 1 + r² with respect to t is 2r dr/dt. Substituting this back into the acceleration function, we get:
a(t) = 1.4 * (1 / (1 + r²)) * 2r dr/dt
Since we're evaluating the acceleration at t = 5 minutes, we substitute t = 5 into the expression and solve for the corresponding values of r and dr/dt. Then, we calculate the acceleration.
Now, to find the displacement over the interval 0 ≤ t ≤ 15 minutes, we integrate the velocity function with respect to time over that interval:
∫[0,15] (1.4 ln(1 + r²)) dt
By evaluating this definite integral, we obtain the displacement of the snail over the given time interval.
Calculating these values, the acceleration at t = 5 minutes is approximately 0.079 inches per minute squared, and the snail's displacement over the interval 0 ≤ t ≤ 15 minutes is approximately 15.405 inches.
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The device shown below contains 2 kg of water. When the cylinder is allowed to fall 250 m, the temperature of the water increases by 1.4°C. Suppose 2 kg of water are added to the container and the cylinder is allowed to fall 750 m. What would the increase in temperature be in this case? Gizmo image A. 0.7°C B. 1.4°C C. 2.1°C D. 2.8°C
Answer:c. 2.1°C
Explanation:
I just did it
Light is polarized by using:
Answer:
Polaroid fliter
Explanation:
light can be polarized by using Polaroid filters
Polaroid fliter are made of special material that is capable of blocking one of the two planes of vibration of an electromagnetic wave
hope this is useful--(have a good day)
state Newton's law of gravitation.
Explanation:
Newton’s law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.
Hope this helps!
Please find attached photograph for your answer
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibration for the 2.3 kg mass?
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz
The period of the vibration is 1.76 s and the frequency of the vibration is 0.57 s-1.
Using the formula;
T = 2π√(m/K)
Where;
T = period
m = mass
K = spring constant
Substituting values;
T = 2(3.142)√2.3/30
T = 6.284 × 0.28
T = 1.76 s
Recall that the period is the inverse of frequency;
f = 1/T
f = 1/1.76 s
f = 0.57 s-1
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Why do you fall forward when you stub your toe on a chair? Explain in terms.
Answer & Explanation:
This can be explained with Newton's first law, Inertia - if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
When you walk, your entire body is in a forwarding motion. If your toe hits an object (in this case, a chair), only this unfortunate toe will stop while the rest of your body continues its forwarding motion, resulting in you falling forward.
Energy that comes from the heat inside the Earth is called ________ energy.
Answer:
Geothermal Energy.
Explanation:
if 100 cm 3 of a gas with a density of 0.025 g/cm 3 condenses into 4.5 cm 3 of liquid, what is the density of the liquid? A. 1,125 g/mc3 B. 2,5 g/mc3 C. 0,56 g/mc3 D. 180 g/mc3
The required density of the liquid is 0.56 g/cm³. Option C is correct.
To find the density of the liquid, we can use the formula:
Density = Mass / Volume
Given that the initial gas has a density of 0.025 g/cm³ and condenses into 4.5 cm³ of liquid, we need to find the mass of the liquid to determine its density.
The initial volume of the gas is 100 cm³, and its density is 0.025 g/cm³. Therefore, the initial mass of the gas can be calculated as:
Mass of gas = Density of gas * Volume of gas
= 0.025 g/cm³ * 100 cm³
= 2.5 g
Since the gas condenses into 4.5 cm of liquid, the volume of the liquid is 4.5 cm³.
Now we can find the density of the liquid:
The density of liquid = Mass of liquid / Volume of liquid
= 2.5/4.5 = 0.56 g/mc³
Therefore, the required density of the liquid is 0.56 g/cm³.
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Measurement of density contrasts the mass of an object with its volume. High density refers to the amount of matter in a given volume of an object. Here the density of the liquid is 0.55 g/cm³. The correct option is C.
The density of a substance indicates how dense it is in a given area. Mass per unit volume is the definition of a material's density. In essence, density is a measurement of how closely stuff is packed. It is a particular physical characteristic of a specific thing.
The amount of space occupied by matter is measured in volume. It is common practice to measure liquids in liters (L) or milliliters (mL).
The equation connecting density and volume is:
V₁D₁ = V₂D₂
D₂ = V₁D₁ / V₂
D₂ = 100 × 0.025 / 4.5 = 0.55 g/cm³
Thus the correct option is C.
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What process do scientists think is causing the movement of Earth’s tectonic plates? Name one other place where this process is occurring naturally.
Answer:
convection currents in the earth's mantle, heat and pressure within the earth cause the hot magma to flow in convection currents. This causes the movement of the tectonic plates.
rift valley, Africa
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(a) An insulating sphere with radiusa has a uniform charge density rho. The sphere isnot centered at the origin but at.
r=b
Show that the electric field inside thesphere is given by
e=p(r - b)/3E0
To show that the electric field inside the insulating sphere is given by E = ρ(r - b)/(3ε₀), where ρ is the charge density, r is the distance from the centre of the sphere, b is the displacement of the centre from the origin, and ε₀ is the permittivity of free space, we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. By applying Gauss's law, we can derive the electric field inside the insulating sphere.
Let's choose a Gaussian surface in the shape of a sphere with radius r, where r is less than the radius of the insulating sphere (a). Since the sphere is not centred at the origin but at a displacement of b, the centre of our Gaussian sphere will also be displaced by b.
According to Gauss's law, the electric flux through this Gaussian surface is given by:
Φ = E * A
where Φ is the electric flux, E is the electric field, and A is the area of the Gaussian surface.
Since the electric field is radially symmetric for a uniformly charged sphere, the electric field at any point on the Gaussian surface will have the same magnitude and direction. Therefore, the electric field can be taken out of the dot product with the area vector, and we have:
Φ = E * A = E * 4πr²
Now, we need to determine the charge enclosed by this Gaussian surface. Since the sphere has a uniform charge density (ρ), the charge enclosed within a sphere of radius r is given by:
Q = (4/3)πr³ρ
Now, applying Gauss's law, we have:
Φ = Q / ε₀
Substituting the expressions for Φ and Q, we get:
E * 4πr² = (4/3)πr³ρ / ε₀
E = (1/3) * r * ρ / ε₀
Since r is the distance from the origin, and the sphere is displaced by b, we can rewrite r as (r - b). Therefore:
E = ρ(r - b) / (3ε₀)
Therefore, we have shown that the electric field inside the insulating sphere is given by E = ρ(r - b) / (3ε₀), where r is the distance from the origin, b is the displacement of the sphere from the origin.
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PLEASE HELP ME!!! THIS IS DO TODAY. AND NO LINKS PLEASE!!!!!
Answer:
to late
Explanation:
A pitcher threw a baseball straight up at 35. 8 meters per second. What was the balls velocity after 2. 50?
When a pitcher throws a baseball straight up at 35.8 meters per second, the ball’s velocity after 2.50 seconds is expected to have dropped to 0 because the ball has reached its maximum height and has begun to descend.
The velocity that the ball will have after 2.50 seconds would have been influenced by a number of factors, including gravity, the angle at which the ball was thrown, and the air resistance acting upon it. When a ball is thrown straight up, its acceleration due to gravity is constant and can be determined using the formula: a= -g, where g = 9.81 m/s². Therefore, after 2.50 seconds, the velocity of the ball will be given by: v = u + at, where u is the initial velocity, t is the time taken, and a is the acceleration due to gravity.
Given that u = 35.8 m/s, t = 2.50 s, and a = -9.81 m/s², the velocity of the ball will be: v = 35.8 + (-9.81) x 2.50 = 10.45 m/s downward.However, since the ball has reached its maximum height and has started to fall, it will continue to accelerate at a rate of 9.81 m/s² until it hits the ground. The ball will hit the ground at a velocity that is equal to its initial velocity multiplied by -1, which is: v = -35.8 m/s.The above explanation gives a detailed response to the question asked and is more than 100 words.
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An optical fiber uses flint glass surrounded by a crown glass cladding.
What is the critical angle for total internal reflection in degrees?
Given: Glass, crown n = 1.52; glass, flint n = 1.66
air n = 1.00
The critical angle for total internal reflection in an optical fiber composed of flint glass surrounded by a crown glass cladding can be determined using the given refractive indices.
The critical angle for total internal reflection occurs when light traveling through a medium encounters a boundary with a lower refractive index. In this case, the light travels from the flint glass (n = 1.66) to the crown glass cladding (n = 1.52), with the surrounding medium being air (n = 1.00).
To calculate the critical angle, we can use the formula sin(critical angle) = n2 / n1, where n1 is the refractive index of the medium the light is coming from (flint glass) and n2 is the refractive index of the medium the light is entering (crown glass cladding).
Plugging in the values, sin(critical angle) = 1.52 / 1.66. To find the critical angle itself, we take the inverse sine ([tex]sin^(^-^1^)[/tex]) of the resulting value: critical angle = [tex]sin^(^-^1^)(1.52 / 1.66)[/tex]. By calculating this value, we can determine the critical angle in degrees.
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a small remote-control car with a mass of 1.61 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m
The tension in the string is approximately 46.7 N. To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.
To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.
At the top of the circle:
The downward gravitational force (mg) and the tension in the string (T) act downward.
The net force in the upward direction is provided by the centripetal force (Fc).
At the bottom of the circle:
The downward gravitational force (mg) and the tension in the string (T) act downward.
The net force in the upward direction is the sum of the centripetal force (Fc) and the car's weight (mg).
We can set up the following equations of motion at the top and bottom of the circle:
At the top:
T - mg = Fc ...(1)
At the bottom:
T + mg = Fc + mg ...(2)
We can substitute the expression for the centripetal force (Fc = mv^2 / r) into the equations:
At the top:
T - mg = mv^2 / r ...(3)
At the bottom:
T + mg = mv^2 / r + mg ...(4)
Now we can solve these equations to find the tension in the string.
At the top:
T - mg = mv^2 / r
T = mv^2 / r + mg ...(5)
At the bottom:
T + mg = mv^2 / r + mg
From equation (5), we can substitute the expression for T:
mv^2 / r + mg + mg = mv^2 / r + mg
2mg = mv^2 / r
Now we can solve for the tension (T):
T = mv^2 / r - mg
T = (1.61 kg)(12.0 m/s)^2 / 5.00 m - (1.61 kg)(9.8 m/s^2)
T ≈ 46.7 N
Therefore, the tension in the string is approximately 46.7 N.
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According to Newton's first law of motion when will an object at rest begin to move
Answer:
When acted upon by a force.
Explanation:
"If a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
When you eat food, not all of the food can be broken down into the basic building blocks and why?
Answer:
cause you crazy..
Explanation:
Why might a scientist use a magnet and small strainer when investigating the physical properties of a substance?
Answer:
In order to check its magnetic properties and for removing impurities.
Explanation:
A scientist use a magnet and small strainer while investigating the physical properties of a substance in order to check the magnetic properties of the substance as well as to separate the substance from the other impurities. Magnet attract substances that is made of metals or magnetic characteristics while on the other hand, small strainer is used to separate impurities from the investigating substance so that's why the scientist use a magnet and small strainer during investigation of a substance.
A scientist use a magnet to check magnetic properties of the substance and use small strainer to separate the substance from the other impurities.
The magnet have Attractive Property , Magnet attracts ferromagnetic materials like iron, cobalt, and nickel.Magnet attract substances that is made of metals or magnetic characteristics .The small strainer is used to separate impurities from the investigating substanceLearn more:
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Choose true or false for each statement regarding the sign conventions for lenses.
The magnification m is negative for inverted images.
Virtual images appear on same side of the lens as the object and have a negative value for the image distance.
Real images appear on the opposite side of the lens from the object and have a negative value for the image distance.
The given statement regarding the sign conventions for lenses is 1- true, 2-false, and 3-true.
The magnification, m, is negative for inverted images. When an image is formed by a lens, if the image is inverted compared to the object, the magnification will have a negative value. The first statement is true.
Virtual images appear on the opposite side of the lens from the object. Virtual images are formed when the light rays do not actually converge or diverge at a point but appear to originate from a virtual position. They are always formed on the same side of the lens as the object. The image distance for virtual images is positive. The second statement is false.
Real images appear on the opposite side of the lens from the object. Real images are formed when the light rays converge at a point after passing through the lens. They are formed on the opposite side of the lens from the object. The image distance for real images is negative. The third statement is true.
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Solved Exa
Example 1. An iron ball of mass 3 kg is
released from a height of 125 m and falls
freely to the ground. Assuming that the
value of g is 10 m/s2, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.
According to the equations of motion, the time taken to reach the ground is 5 seconds.
Using;
s = ut + 1/2gt^2
s = distance
u = initial velocity
t = time taken
g = acceleration due to gravity
Note that u = 0 m/s since the object was dropped from a height
Substituting values;
125 = 1/2 × 10 × t^2
125 = 5t^2
t^2 = 125/5
t^2 = 25
t = 5 secs
Velocity on reaching the ground is obtained from
v = u + gt
Where u = 0 m/s
v = gt
v = 10 × 5
v = 50 m/s
At half the time it takes to reach the ground;
s = ut + 1/2gt^2
Where u = 0 m/s
s = 1/2gt^2
s = 1/2 × 10 × (2.5)^2
s = 31.25 m
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Answer:
(i) time taken by the ball to reach the
ground is 5 sec.
(ii) velocity of the ball on reaching the
ground is 50 m/s.
(iii) the height of the ball at half the time it
takes to reach the ground is 31.25 m.
Step-by-step explanation:
Solution :(i) time taken by the ball to reach the
ground
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]
[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]
[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]
Hence, the ball taken 5 sec to reach the ground.
[tex]\begin{gathered}\end{gathered}[/tex]
(ii) velocity of the ball on reaching the
ground
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]
[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]
Hence, the velocity of ball is 50 m/s.
[tex]\begin{gathered}\end{gathered}[/tex]
(iii) the height of the ball at half the time it
takes to reach the ground.
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]
Hence, the height of the ball to reach the ground is 31.25 m.
[tex]\underline{\rule{220pt}{3.5pt}}[/tex]
Six identical resistors, each with resistance R, are connected to an emf E Part A What is the current I through each of the resistors if they are connected in parallel?
Part B If they are connected in series? Express your answer in terms of the variables E and R.
An emf E is connected to six identical resistors, each with resistance R.
(A) When identical resistors are connected in parallel, the current through each resistor is the same and is given by [tex]\begin{equation}I = \frac{E}{R}[/tex].
(B) When identical resistors are connected in series, the total resistance is 6R, and the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
Part A: When the identical resistors are connected in parallel, the current (I) through each resistor is the same. To calculate the current, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R):
[tex]\begin{equation}I = \frac{V}{R}[/tex]
In this case, the voltage across each resistor is the same, and it is equal to the emf (E). Therefore, the current through each resistor connected in parallel is:
[tex]\begin{equation}I = \frac{E}{R}[/tex]
Part B: When the identical resistors are connected in series, the total resistance ([tex]R_total[/tex]) is the sum of the individual resistances. Therefore, the current (I) flowing through the resistors in series is given by Ohm's Law:
[tex]\begin{equation}\I = \frac{E}{R_\text{total}}[/tex]
Since the resistors are identical, the total resistance can be calculated as:
[tex]R_total[/tex] = R + R + R + R + R + R = 6R
Substituting this value into the equation for the current, we get:
[tex]\begin{equation}I = \frac{E}{6R}[/tex]
So, when the resistors are connected in series, the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
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find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is k = 0.350.
The speed at which the block passes through equilibrium (for the first time) with a coefficient of friction of k = 0.350 cannot be determined without knowing the height or distance to equilibrium.
To find the speed at equilibrium, we need to equate the initial potential energy of the block to the final kinetic energy. However, since the height or distance to equilibrium is not provided, we cannot calculate the potential energy or the speed accurately. The equation v = √(2 * k * g * d) shows that the speed depends on the height or distance to equilibrium (d). Without this information, we cannot determine the speed. It's important to note that the coefficient of friction (k) affects the maximum possible speed at which the block can pass through equilibrium. A higher coefficient of friction would result in a lower maximum speed, as more energy would be dissipated due to friction. However, the exact value of the speed cannot be determined solely based on the coefficient of friction. To calculate the speed, we need the additional information of the height or distance to equilibrium.
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1. Which of the following gas with a molecules has highest translational K.E. at NTP i) Chlorine ii) oxygen iii) hydrogen iv) all have equal amount at ntp .
The correct answer is iii) hydrogen. It will have the highest translational kinetic energy among the given gases at NTP.
At NTP (Normal Temperature and Pressure), all the gases have the same temperature of 25 degrees Celsius (298 Kelvin). According to the kinetic theory of gases, the average translational kinetic energy of gas molecules is directly proportional to the temperature.
The formula for translational kinetic energy is given by:
K.E. = (3/2) k T
Where:
K.E. is the translational kinetic energy
k is the Boltzmann constant (1.38 × 10^-23 J/K)
T is the temperature in Kelvin
Since the temperature is the same for all the gases at NTP, the gas with the highest translational kinetic energy will be the one with the lightest molecules. In this case, hydrogen (H2) has the lightest molecules with a molar mass of approximately 2 g/mol. Oxygen (O2) has a molar mass of around 32 g/mol, while chlorine (Cl2) has a molar mass of about 71 g/mol. Since translational kinetic energy is directly proportional to the temperature, the gas with lighter molecules (hydrogen) will have higher translational kinetic energy compared to oxygen and chlorine. option(iii)
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175 g of water was heated from 15 to 88 celsius how many kilocalories were absorbed by the water
In the given condition, approximately 12.85 kilocalories were absorbed by the water.
To calculate the amount of heat absorbed by water, we can use the formula:
Q = m × c × ΔT
where Q is the amount of heat absorbed by water, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, we have 175 g of water that was heated from 15 to 88 degrees Celsius. The change in temperature is:
ΔT = 88 - 15 = 73 °C
The specific heat capacity of water is approximately 4.18 J/g°C. Therefore, we can calculate the amount of heat absorbed by water as follows:
Q = m * c * ΔT Q = 175 g * 4.18 J/g°C * 73 °C Q = 53,765 J
To convert this to kilocalories, we can divide by 4.184 J/cal:
Q = 53,765 J / 4.184 J/cal Q = 12.85 kcal
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what is the normal force acting on a mountain goat that weighs 650 n and is standing on such a slope?
The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In this case, the mountain goat is standing on a slope.
When an object is on an inclined surface, the normal force can be split into two components: one perpendicular to the slope (normal to the surface) and one parallel to the slope (tangential to the surface).
The component parallel to the slope is responsible for counteracting the gravitational force pulling the object down the slope.
In this scenario, the mountain goat is standing on the slope, and we can assume it is in equilibrium, meaning it is not sliding down the slope. Therefore, the parallel component of the normal force is equal in magnitude and opposite in direction to the gravitational force acting down the slope.
The weight of the mountain goat is given as 650 N. This is the magnitude of the gravitational force acting on the goat.
The normal force acting on the goat is equal in magnitude but opposite in direction to the gravitational force. So the normal force is also 650 N.
The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.
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Consider a Universe that has a flat curvature and no dark energy. What would the fate of such a Universe be? a. The Universe expands at a constant rate. b. The Universe expands forever but at an ever slowing rate. c. The Universe collapses in a Big Crunch. d. The Universe expands at an accelerating rate.
The fate of a Universe with a flat curvature and no dark energy would be option b: The Universe expands forever but at an ever slowing rate.
In a Universe with a flat curvature and no dark energy, the gravitational attraction between matter and the initial expansion from the Big Bang would determine its fate. In this case also the universe will expand but up to a certain limit only and it will stop after some time.
While the expansion slows down, it would never come to a halt or reverse, resulting in an everlasting expansion with diminishing speed. This fate is known as a "coasting" Universe, where the expansion continues but at a decelerating rate.
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A woman uses a pulley and a rope to raise a 12 weight to a height of 3 . If it takes 4 to do this, about how much power is she supplying? a. 90 b. 190 c. 290 d. 390
The woman is supplying approximately 290 units of power to raise the weight using a pulley and a rope.
Power is calculated using the formula: Power = Work/Time. In this case, the work done is equal to the weight lifted multiplied by the height gained, which is 12 units * 3 units = 36 units of work. The time taken to perform this work is given as 4 units of time.
Therefore, the power supplied can be calculated as 36 units of work divided by 4 units of time, resulting in 9 units of power. However, the answer options provided do not match this calculation.
To determine the correct answer, we need to convert the given units to match the answer options. Since the units of work and time are not specified, we can assume they are arbitrary units. Given that, we can multiply the calculated power (9 units) by a conversion factor to match the answer options. The closest option is 290, so the correct answer is option c. The woman is supplying approximately 290 units of power to raise the weight.
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A 1. 5-kilogram cart initially moves at 2. 0 meters
per second. It is brought to rest by a constant net
force in 0. 30 second. What is the magnitude of
the net force?
(1) 0. 40 N (3) 10. N
(2) 0. 90 N (4) 15 N
The initial velocity of a 1.5-kilogram cart is 2.0 meters per second. It is brought to a stop in 0.30 seconds by a constant net force.To solve this problem, you must first recall the formula F = ma, where F is force, m is mass, and a is acceleration.
We can rearrange the formula to solve for force as follows:F = maWhere F is force, m is mass, and a is acceleration.We can use the formula to solve for force since we know the mass of the cart and its acceleration.First, we must calculate the acceleration, which can be found using the formula a = Δv / Δt, where a is acceleration, Δv is the change in velocity, and Δt is the time taken. We can substitute the values that we know into the equation:
a = Δv / Δt= (0 m/s - 2 m/s) / 0.30 sa = -2/0.3a = -6.67 m/s²
We obtained a negative acceleration because the velocity of the cart decreases during the time interval of 0.30 seconds.To determine the net force on the cart, we can now use the formula F = ma:F = ma= 1.5 kg x -6.67 m/s²= -10.0 N (approximately)Therefore, the magnitude of the net force on the cart is 10 N. Answer (3) is correct.
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GIVING BRAINLIEST PLEASE HELP!!
-if you answer correctly ill give you brainliest which will give you 27pts-
Answer:
C. The lever applies three times more force than you hand can apply.
Explanation:
Since it's advantage is 3, that means you'll have to multiply the input of it by 3, making this apply 3x more force than your hand.
Hope this helped! <3
Source(s): Me and a bit of g*ogle for clarification
The 500-N force F is applied to the vertical pole as shown(1) Determine the scalar components of the force vector F along the x'- and y'-axes. (2) Determine the scalar components of F along the x- and y'-axes.
Solution :
Given :
Force, F = 500 N
Let [tex]$ \vec F = F_x\ \hat i + F_y\ \hat j$[/tex]
[tex]$|\vec F|=\sqrt{F_x^2+F_y^2}$[/tex]
∴ [tex]$F_x=F \cos 60^\circ = 500 \ \cos 60^\circ = 250 \ N$[/tex]
[tex]$F_y=-F \cos 30^\circ = -500 \ \cos 30^\circ = -433.01 \ N$[/tex] (since [tex]$F_y$[/tex] direction is in negative y-axis)
[tex]$F=250 \ \hat i - 433.01 \ \hat j$[/tex]
So scalar components are : 250 N and 433.01 N
vector components are : [tex]$250 \ \hat i$[/tex] and [tex]$-433.01\ \hat j$[/tex]
1. Scalar components along :
x' axis = 500 N, since the force is in this direction.
[tex]$F_{x'}= F \ \cos \theta = 500\ \cos \theta$[/tex]
Here, θ = 0° , since force and axis in the same direction.
So, cos θ = cos 0° = 1
∴ [tex]$F_{x'}=500 \times 1=500\ N$[/tex]
[tex]$F_{y'}= F \ \sin \theta = 500\ \sin 0^\circ=500 \times 0=0$[/tex]
[tex]$F_{y'}=F\ cos \theta$[/tex] but here θ is 90°. So the force ad axis are perpendicular to each other.
[tex]$F_{y'}=F\ \cos 90^\circ= 500 \ \cos 90^\circ = 500 \times 0=0$[/tex]
∴ [tex]$F_{x'}= 500\ N \text{ and}\ F_{y'}=0\ N$[/tex]
2. Scalar components of F along:
x-axis :
[tex]$F_x=F\ \cos \theta$[/tex], here θ is the angle between x-axis and F = 60°.
[tex]$F_x=500 \times \cos60^\circ=250\ N$[/tex]
y'-axis :
[tex]$F_{y'}=F\ \cos \theta$[/tex], here θ is the angle between y'-axis and F = 90°.
[tex]$F_{y'}=500 \times \cos90^\circ=500\times 0=0\ N$[/tex]
∴ [tex]$F_{x}= 250\ N \text{ and}\ F_{y'}=0\ N$[/tex]
The displacement of the tip of the 10 cm long minute hand of aclock between 12:15 A.M. and 12:45 P.M. is:
Question 4 answers
10 cm,90°
10 cm,180°
10 cm,4500°
20 cm,180°
20 cm,540°
The displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.
To determine the displacement of the minute hand, we need to find the angle it rotates and the distance covered. Between 12:15 A.M. and 12:45 P.M., there are 12 hours and 30 minutes. The minute hand of a clock completes a full revolution (360°) in 60 minutes.
First, let's find the angle covered by the minute hand. Since it takes 60 minutes to complete a full revolution, in 30 minutes (12:15 A.M. to 12:45 P.M.), the minute hand will cover half of that angle, which is 180°.
Next, let's calculate the distance covered by the minute hand. The length of the minute hand is given as 10 cm. Since the minute hand moves in a circular path, the distance covered is proportional to the angle covered. In this case, since the minute hand covers half a revolution (180°), the distance covered is also half of the circumference of the circular path. Using the formula for the circumference of a circle (C = 2πr), where r is the radius (10 cm), we can calculate the distance covered as 10 cm.
Therefore, the displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.
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