Answer:
A. harmful substances in the air, water, or soil
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On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 20,000 N. a. Show that the weight of the displaced air is 20,000 N. b. Show that the volume of the displaced air is 1700 m3 .
Explanation:
Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.
Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.
Weight of the air displaced = density of air × volume
The density of air at 1 atm pressure and 20º C is 1.2 kg/m³
the volume V = 20,000/(1.2×9.8) = 1700 m³
Which action can knowing the "Three Rs lead a person to do?
A) resolving conflicts
B) helping one's family create a disaster plan
C)staying safe in a natural disaster
D)protecting the environment
Answer:
d
Explanation:
In a home stereo system, low sound frequencies are handled by large "woofer" speakers, and high frequencies by smaller "tweeter" speakers. For the best sound reproduction, low-frequency currents from the amplifier should not reach the tweeter. One way to do this is to place a capacitor in series with the 9.0 Ω resistance of the tweeter; one then has an RLC circuit with no inductor L (that is, an RLC circuit with L = 0).
What value of C should be chosen so that the current through the tweeter at 200 Hz is half its value at very high frequencies? Express your answer with the appropriate units.
Answer:
C = 2.9 10⁻⁵ F = 29 μF
Explanation:
In this exercise we must use that the voltage is
V = i X
i = V/X
where X is the impedance of the system
in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is
X = [tex]\sqrt{R^2 + ( wL - \frac{1}{wC})^2 }[/tex]
tells us to take inductance L = 0.
The angular velocity is
w = 2π f
the current is required to be half the current at high frequency.
Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small
[tex]\frac{1}{wC}[/tex] →0 when w → ∞
therefore in this frequency regime
X₀ = [tex]\sqrt{R^2 + ( \frac{1}{2\pi 2 10^4 C} )^2 } = R \sqrt{ 1+ \frac{8 \ 10^{-10} }{RC} }[/tex]
the very small fraction for which we can despise it
X₀ = R
to halve the current at f = 200 H, from equation 1 we obtain
X = 2X₀
let's write the two equations of inductance
X₀ = R w → ∞
X= 2X₀ = [tex]\sqrt{R^2 +( \frac{1}{wC} )^2 }[/tex] w = 2π 200
we solve the system
2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }
4 R² = R² + 1 / (wC) ²
1 / (wC) ² = 3 R²
w C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{R}[/tex]
C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{wR}[/tex]
let's calculate
C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{2\pi \ 200 \ 9}[/tex]
C = 2.9 10⁻⁵ F
C = 29 μF
A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are µs = 0.50 and µk = 0.40. For P = 3.6 lb,
Determine:
(a) Whether slipping occurs between the belt and either cylinder,
(b) The angular acceleration of each cylinder.
Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.
Given the following data:
Mass A = 5 lb to kg = 2.27 kg.
Mass B = 20 lb to kg = 9.02 kg.
Force = 3.6 lb to N = 16.02 Newton.
How to calculate angular acceleration.In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.
For cylinder A:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]
For cylinder B:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]
For the belt, we have
[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]
Also, we would determine the angular acceleration of cylinder B:
[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]
Next, we would calculate the forces acting on the cylinders:
[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]
[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]
Next, we would determine the force of static friction:
[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]
From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.
Read more on angular acceleration here: https://brainly.com/question/6860269
an object is acted by force of 22 newtons to the right and a force of 13 newtons to left
Answer:
Explanation:
I'm going to guess that you want the net force.
These two forces are acting in opposite directions. Therefore the forces are subtracted in effect.
F = F1 - F2
F1 = 22N to the right
F2 = 13N to the left
F = 22 - 13 = 9 N to the right.
Specifying the direction is very important. Forces do have directions and you must specify what that is. Otherwise, the question should be marked incorrect.
the pencil has a shadow. is the pencil opaque or transparent?
The pencil is clearly an opaque object because we cannot see through it and therefore forms a shadow.
sample of pure boron contains only isotope X and isotope Y.
A nucleus of X has more mass than a nucleus of Y.
[o[4].[4] The sample is ionised, producing ions each with a charge of +1.6 x 10°C.
The specific charge of an ion of X is 8.7 x 10°C kg".
Calculate the mass of an ton of X.
[1 mark]
h
mass of ion = kg
[4].[2] Determine the number of nucleons in a nucleus of X.
mass of a nucleon = 1.7 x 1077 kg
[2 marks]
h
number of nucleons =
[o[4].[3] Compare the nuclear compositions of X and Y.
[2 marks]
[o[4].[4] lons of Y have the same charge as ions of X.
State and explain how the specific charge of an ion of X compares with that of an
ion of Y.
[2 marks]
Answer:
[tex]1.84[/tex]×[tex]10^{-26}[/tex]
Explanation:
specific charge = [tex]\frac{charge}{mass}[/tex] so by rearranging for mass we get
mass= [tex]\frac{charge}{sepcifc charge}[/tex]
[tex]\frac{1.6×10^{-26} }{1.7×10^{-27} }[/tex] = answer in kg
Three 5 Ohm resistors are connected in series to a 10 Volt power supply. What is the current through each resistor?
Answer74.3
:
Explanation:
a disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?
Answer:
a) α = 1.875 [tex]\frac{rad}{s^{2} }[/tex]
b) t = 8 s
Explanation:
Given:
ω1 = 0 [tex]\frac{rad}{s}[/tex]
ω2 = 15 [tex]\frac{rad}{s}[/tex]
theta (angular displacement) = 60 rad
*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*
a) α = ?
(ω2)^2 = (ω1)^2 + 2α(theta)
[tex]15^{2}[/tex] = [tex]0^{2}[/tex] + 2(α)(60)
225 = 120α
α = 1.875 [tex]\frac{rad}{s^{2} }[/tex]
b)
α = (ω2-ω1)/t
t = (ω2-ω1)/α = (15-0)/1.875 = 8
t = 8 s
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 60.0 N is exerted and the astronaut's acceleration is measured to be 0.870 m/s2. (a) Calculate her mass (in kg). kg (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut's acceleration. Propose a method in which recoil of the vehicle is avoided.
Answer:
a) m = 69.0 kg
b) release some gas in the opposite direction to the astronaut's movement
Explanation:
a) Let's use Newton's second law
F = m a
m = F / a
m = 60.0 / 0.870
m = 69.0 kg
b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved
p₀ = p_f
m v = M v '
v ’= [tex]\frac{m}{M} \ v[/tex]
so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.
One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.
Please help me someone !
Answer:
The object is moving at constant speed.
Explanation:
The spaces between the dots are equal.
TIME REMAINING
15:56:15
A plant root is an example of
Type here to search
Answer:
h
Explanation:
What is the equation for frequency?
a. number of cycles +unit of time
b. number of cycles - unit of time
c. number of cycles ×unit of time
d. number of cycles/ unit of time
Answer:
d
Explanation:
An electromagnetic wave has a frequency of 5.0 x 1014 Hz. What is the
wavelength of the wave? Use the equation 2 = and 3.0 x 108 m/s for the
speed of light.
A. 1.7 x 10-8 m
О
B. 6.0 x 1022 m
O C. 6.0 x 10-7 m
O D. 1.7 x 105 m
Answer:
c
Explanation:
wavelength = speed of light/ frequency
= (3x 10^8 m/s)/(5.0 x 10^14 Hz)
= 6.0 x 10^-7 m
A cylindrical diving bell is open at the bottom and closed at the top, and is 5m tall. The bell is open to atmospheric air until it is placed in the water, but the bell remains upright (open end facing down, closed end facing up). The pressure of the air inside the bell will naturally increase by 105 Pa for every 10m of depth the bell descends within the water. Assume the temperature of the air remains constant for this process, and that the air can be approximated as an ideal gas
Required:
a. If the bell is lowered 40 meters below the surface, how many meters of air space are left inside the bell?
b. Explain why water doesn't completely flood the bell as it enters the water V (m).
Answer:
a) y = 0.35 m, b) hydrostatic balance
Explanation:
a) For this fluid mechanics exercise, let's use that the pressure at a given level is the same, let's set a level on the bell shape.
The pressure inside is
P_interior = P₀ + ρ g h ’
The pressure outside
P_exterior = Pₐ + ρ g h
as the point is at the same level the pressures are equal
P_interior = P_exterior
P₀ + ρ g h ’= Pₐ + ρ g h
h ’= (Pₐ- P₀) + ρ g h
To calculate P₀ they indicate that the pressure increases 10⁵ Pa for every 10 m, we use a direct rule of proportions or rule of three
P₀ = 10⁵ (40 + h ’) / 10 = 4 10⁵ + h’ 10⁴
the positive sign is because the water inside the hood also increases the air pressure.
we substitute
(4 10⁵ + h’ 10⁴) + ρ g h’ = Pₐ + ρ g h
h’ (ρ g + 10⁴) = Pₐ - 4 10⁵ + ρ h h
h’ (1000 9.8 + 10⁴) = (1 10⁵ -4 10⁵) + 1000 9.8 40
h' (1.98 10⁴) = -3 105 + 3.92 10⁵
h’ = [tex]- \frac{0.92 \ 10^5 }{1.98 \ 10^4 }[/tex]
h ’= -4.65 m
as the hood is only 5 m high, the free air space is
Y = 5 - 4.65
y = 0.35 m
it is very little free space
B) The pressure outside and inside the hood is the same, the water rises inside the hood until the pressures equalize and at this point the force is equal and in the opposite direction, which is why the system is in hydrostatic balance.
How large a force F is needed in the figure to pull out the 6.0 kg block with an acceleration of 1.50m/s^2 if the coefficient of friction at its surfaces is 0.40?
Answer:
32.52 N
Explanation:
F-F' = ma.................. Equation 1
Where F = Force needed to pull the block, m = mass of the block, a = acceleration of the block, F' = Frictional force of the surface acting on the blcok
F = ma+F'
But,
F' = μmg
Where g = acceleration due to gravity, μ = coefficient of static friction
F = ma+μmg.................... Equation 2
Given: m = 6.0 kg, a = 1.5 m/s², μ = 0.40
Constant: g = 9.8 m/s²
Substitute these values into equation 2
F = 6(1.5)+6(9.8)(0.4)
F = 9+23.52
F = 32.52 N
how would a small bar magnet be oriented when placed at position x ? option c is wrong.
Answer:
B
Explanation:
Heat transfer in liquids or gases that happens due to currents of hot and cold is called
Entropy
Conduction
Convection
Radiation
Answer:
Convection
Explanation:
three types of heat transfer
Heat is transfered via solid material (conduction), liquids and gases (convection), and electromagnetical waves (radiation).
When a solid uniform sphere is spinning about an axis of rotation through its center, its rotational kinetic energy is K and moment of inertia I = ⅖ MR2. A second solid sphere having twice the mass and twice the diameter of the first one is spinning about an axis through its center and has a twice the angular velocity of the first sphere. The rotational Kinetic energy of the second sphere is:
8K
4K
32K
2K
K
Answer:
uh.
Explanation:
A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
Answer:
a) v = 3.116 m / s, b) μ = 1.65 10⁻²
Explanation:
a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = M v₀
final instant. After the crash
p_f = (M + m) v
the moment is preserved
M v₀ = (M + m) v
v = [tex]\frac{M}{M+m} \ v_o[/tex]
let's calculate
v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]
v = 3.116 m / s
b) for this part let's use the relationship between work and kinetic energy
W = ΔK
as the body has its final kinetic energy is zero
the work of the friction forces is
W = - fr x
the negative sign is because the friction forces always oppose the movement
let's write Newton's second law
Y axis
N - W_sled -W_person = 0
N = mg + M g
N = (m + M) g
X axis
fr = ma
the friction force has the expression
fr = μ N
fr = μ g (m + M)
we substitute
- μg (m + M) x = 0- ½ (m + M) v²
μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]
let's calculate
μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]
μ = 0.0165
μ = 1.65 10⁻²
does altitude has an effect on weight, yes or no
Answer: yes
Explanation:
Weight is the gravitational force experienced on a body. If you move up to higher altitudes, the distance between you and earth increases. ... Yes, weight drops as you go up in altitude (because of diminishing gravity), though your mass remains the same. However, the effect is not huge.
A bowling ball and a baseball both roll across your foot at the same speed. The bowling ball hurts much more.
Which law of motion is this?
Answer:
Newtons second law
Explanation:
Depends on mass
Answer:
2nd law
Explanation:
:))))))))))))))
stainless steel, tell us about its properties and what should be taken into account when using it?
A 1-kg ball is 12 m above the ground, with an initial velocity = 0 m/s.
Use the following formulas
[ KE = 1/2 xm x V2]
[P.E = mxgxh] . g =9.8 m/s2
[ Mechanical energy = K.E + P.E]
Answer:
M = 117.6 J
Explanation:
Given that,
The mass of a ball, m = 1kg
The height of the ball, h = 12 m
At point A, its initial velocity, v = 0
The mechanical energy is the sum of kinetic and potential energy such that,
[tex]M=\dfrac{1}{2}mv^2+mgh\\\\M=0+mgh\\\\M=1\times 9.8\times 12\\\\M=117.6\ J[/tex]
So, the mechanical energy is equal to 117.6 J.
Sonya hears water dripping from the eaves of the house onto a porch roof. She counts 30 drops in 1.0 min.
A rod that is 96.0 cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 19.1 cm. An object is in air on the long axis of the rod 19.9 cm from the end that has the 19.1-cm radius.
(a) Find the image distance due to refraction at the 19.1-cm radius surface.
(b) Find the position of the final image due to refraction at both surfaces.
(c) Is the final image real or virtual?
In a solar system far, far away the sun's intensity is 400 W/m2 for an inner planet located a distance R away. What is the sun's intensity for an outer planet (in W/m2) located at a distance of 4 R from the Sun
Answer:
I₂ = 25 W / m²
Explanation:
Intensity is defined as the relationship between power and area
I = P / A
the power emitted by the sun is constant
P = I A
for the two points of interest
I₁ A₁ = I₂ A₂
energy is distributed on the surface of a sphere
A = 4π R²
I₁ R₁² = I₂ R₂²
I₂ = [tex]\frac{R_1^2}{R_2^2} \ I_1[/tex]
let's calculate
I₂ = [tex]\frac{R^2}{(4R)^2} \ 400[/tex]
I₂ = 25 W / m²
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Use the scenario to answer the question.
An astronomer discovers a new galaxy using a telescope. The astronomer wants to investigate how the galaxy is moving relative to the Milky Way galaxy.
In one or two sentences, make a hypothesis about the movement of the galaxy and explain at least one way to test the hypothesis.
Answer:
The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you
Explanation:
The hypothesis about the movement of the galaxy is that galaxies are moving far from each other continuously.
What is the milky way galaxy?The milky way galaxy is a galaxy that contains over a hundred billion stars and it also includes our solar system. Its name describes its appearance when viewed from the earth. All the individual stars in the whole sky are a portion of the Milky Way Galaxy, the term "Milky Way" is because of the band of light.
The astronomer has discovered a new galaxy which means our universe is continuously expanding. This is because the universe encloses everything that exists.
Galaxies are moving in space and since the universe space is continuously expanding so the galaxies continuously move from each other. The farther the galaxy is from the milky way which is an observable part, the faster will be moving the galaxy and the closer the galaxy is to the milky way, the slower will be movement of the galaxy.
Learn more about the milky way galaxy, Here:
https://brainly.com/question/2905713
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Humans impact the Earth in good AND bad ways.
A) True
B) False
Answer:
True
Explanation:
yes we can see that we are helping animals but we create pollution which is very bad
Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed v0(v < vesc). Express in terms of v0, rE, ME, and G.
Initially, the energies are:
[tex]U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\ =K_{i}=\frac{1}{2} m v_{0}^{2}[/tex]
At final point, the energies are:
[tex]U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\ K_{f}=\frac{1}{2} m(0)^{2}=0[/tex]
Using conservation law of energy,
[tex]-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\ -\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\ \frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\ \frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}[/tex]
The equation is further simplified as:
[tex]r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\ h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\ &=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\ & h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}[/tex]