The car accelerates with magnitude a such that
31.3 m/s = 25.6 m/s + a (5.4 s)
→ a = (31.3 m/s - 25.6 m/s) / (5.4 s) ≈ 1.056 m/s²
Then the applied force has a magnitude F of
F = (1430 kg) a ≈ 1500 N
The current theory of the structure of the
Earth, called plate tectonics, tells us that the
continents are in constant motion.
Assume that the North American continent
can be represented by a slab of rock 5200 km
on a side and 35 km deep and that the rock
has an average mass density of 2700 kg/m².
The continent is moving at the rate of about
3.8 cm/year.
What is the mass of the continent?
Answer in units of kg.
Answer:
pt 1: [tex]m=1.66698*10^{21} kg[/tex]
Pt 2: [tex]KE=1212.23531 J[/tex]
Explanation:
Information Given: (p = density)
l = 5200km d = 35km p = 2700kg/[tex]m^{2}[/tex]
Part 1: Mass
Find volume [tex]V=(l)^2(d)[/tex][tex]V=(4.2*10^6)^2(35*10^3)[/tex][tex]V=61.74*10^{16}[/tex]Find Mass[tex]m=Vp[/tex][tex]m=(61.74*10^{16})(2700)[/tex][tex]m=1.66698*10^{21}[/tex]Part 2: Kinetic Energy
[tex]v=\frac{3.8cm}{yr}*\frac{m}{100cm}*\frac{yr}{365d}*\frac{d}{24hr}*\frac{hr}{3600s}[/tex][tex]v=1.20497*10^{-9}[/tex][tex]KE=\frac{1}{2}mv^2[/tex]
[tex]KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2[/tex]
[tex]KE=1212.23531 J[/tex]
Part 3: Jogger Speed
set up, because I don't have the mass :(
Information given:
[tex]KE_{jogger}[/tex]
[tex]KE=\frac{1}{2}mv^2[/tex][tex]v_{jogger} =\sqrt{\frac{2KE}{m_{jogger} } }[/tex]Input the valuesHope it helps :)
The corect phase sequence shown
Gas, Liguid, Solid
Lqud, Gas, Solid
Sold, Liguid, Gas
Gas, Solid, Liquid
above i
Answer:
Explanation:
gas, liquid, soild
liquid, Gas, solid
Gas, Solid, liquid
A 500 kg car is moving at 30 m/s. The driver sees a barrier ahead. If the car takes 100 m to come to rest, what is the magnitude of the force necessary to stop the car?
How do you solve this question?
Answer:
F = 2250 [N]
Explanation:
In order to solve this problem, we must first use the following equation of kinematics.
[tex]v_{f}^{2} =v_{o}^{2}-2*a*x[/tex]
where:
Vf = final velocity = 0 (come to rest)
Vo = initial velocity = 30 [m/s]
a = acceleration or desaceleration [m/s²]
x = distance = 100 [m]
[tex](0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}][/tex]
Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force [N]
m = mass = 500 [kg]
a = acceleration = 4.5 [m/s²]
[tex]F = 500*4.5\\F = 2250 [N][/tex]
If an ocean wave passes a stationary pointevery 4 s and has a velocity of 7 m/s, what isthe wavelength of the wave?Answer in units of m.
Answer:
28mExplanation:
Step one:
given data
period T= 4seconds
velocity v= 7m/s
wave lenght λ=?
Step two:
we know that f=1/T
the expression relating period and wave lenght is
v=λ/T
λ=v*T
λ=7*4
λ=28m
The wavelength of the wave is 28m
friction reduces air resistance?
Answer:no
Explanation:
No.....................................
Which is the largest gas that occurs in our atmosphere?
Helium
Nitrogen
Other Gases
Oxygen
Answer:
OXYGEN
Explanation:brainlyist me
Answer:
Nitrogen
Explanation:
Oxygen is second
A 1.00-m3 object floats in water with 30.0% of its volume above the waterline. What does the object weigh out of the water?
Answer:
Object's weight = 6,839.42 N
Explanation:
Given
Above waterline = 30%
Volume of object = 1m^3
Required
Determine the weight of the object
First, we need to calculate its Mass
Mass = Density of Water * Volume of object in water
Density of water = 997kg/m³
If 30% is above waterline, then 70% is in water.
So:
Mass = Density of Water * Volume of object in water
Mass = 997kg/m³ * 70%m³
Mass = 997kg * 70%
Mass = 697.9 kg
The object weight sis then calculated as thus:
Weight = Mass * Acceleration of gravity
Weight = 697.9 kg * 9.8m/s²
Weight = 6 839.42 N
. What is the barycenter of the Moon and Earth?
Answer:
About 1000 miles.
Explanation:
How does reflection differ from refraction and diffraction?
Reflection is the only process in which the wave does not continue moving forward.
Reflection is the only process in which the wave slows down.
Reflection is the only process that involves a change in the wave.
Reflection is the only process that changes the direction of a wave.
Answer: Reflection is the only process in which the wave does not continue moving forward.
Explanation:
Reflection is a process in which the direction of the wave changes when it is exposed to a bounce off barrier. Refraction can be defined as the change in the direction of the wave when the wave passes through one medium to another. Diffraction is a process in which the direction of the wave changes when the wave passes through a particular opening near the barrier.
Answer:
Reflection is the only process in which the wave does not continue moving forward.
Explanation:
What does the phrase “constant velocity” indicate?
a. zero distance
b. zero acceleration
c. constant acceleration
d. deceleration
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm.
Complete Question
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. The student has 70-cm-long arms
Answer:
The value is [tex]w__{rpm} } = 29.17 \ rpm[/tex]
Explanation:
From the question we are told
The distance from the handle to the bottom of the bucket is [tex]d = 35 \ cm = 0.35 \ m[/tex]
The length of the students arm is L = 70 cm = 0.70 m
Generally the acceleration due to gravity experienced by the bucket of water is mathematically represented as
[tex]g = w^2 * r[/tex]
Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as
[tex]r = L + d[/tex]
So
[tex]g = w^2 * ( L + d )[/tex]
= > [tex]w = \sqrt{\frac{g }{ L + d } }[/tex]
= > [tex]w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }[/tex]
= > [tex]w = 3.055 \ rad/s[/tex]
Generally the angular speed in revolution per minute is mathematically represented as
[tex]w__{rpm} } = \frac{w * 60 }{2 \pi }[/tex]
=> [tex]w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }[/tex]
=> [tex]w__{rpm} } = 29.17 \ rpm[/tex]
pplzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helppppp
Answer:
D. The airplane will turn towards the east
Explanation:
If the airplane is thrown straight towards the north, the window which is moving from left (west) to right (east) the wind will knock the plane towards the right (east) since thats the way it is blowing.
PLEASE HELP!
I don't even know what Science is I'm so dumb lol XD
Answer:
C
Explanation:
Sled A has more potential energy because it's mass is 100 kg, and it is higher up than Sled B. The more high up the sled is and the lighter it is, the faster it gets, it creates more and more potential energy.
Hope this helps!
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the edge of the disc is ?m/s2.
Answer:
Centripetal acceleration = 83.77m/s²
Explanation:
Given the following data;
Radius, r = 0.13m
Velocity, v = 3.3m/s
To find centripetal acceleration;
Centripetal acceleration is given by the formula;
[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]
Substituting into the equation, we have;
[tex] Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}[/tex]
[tex] Centripetal \; acceleration, a = \frac {10.89}{0.13}[/tex]
Centripetal acceleration = 83.77m/s²
Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s².
Answer:
the answer is 84
Explanation:
In the past, Africa used to be further away from Europe than it is now
(shown below). What could explain why Africa is closer to Europe now than
it was before? *
Answer: Plates shifting
Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).
Answer:
nvudbwasivnjlscv bwbfvsz
Explanation:
What is the error in this representation of the steps involved in gene therapy?
Answer:
a
Explanation:
A Boeing 737—a small, short-range jet with a mass of 51,000 kg— sits at rest at the start of a runway. The pilot turns the pair of jet engines to full throttle, and the thrust accelerates the plane down the runway. After traveling 940 m, the plane reaches its takeoff speed of 70 m/s and leaves the ground. What is the thrust of each engine?
Answer:
67000N
Explanation:
We solve for the acceleration using the the 3rd constant-acceleration equation.
(Vx)f² = (Vx)i² + 2ax∆x
We have the displacement to be
∆x = Xf - Xi = 940m
Vx = 70m/s
The acceleration = (70m/s)²/2(940m)
= 4900/1880
= 2.61m/s²
From isaac newton's second law,
51000kg x 2.61m/s²
= 133,000N
The engines thrust is half of this value
Therefore thrust = 67000N or 67kN
Mike rides his horse with a constant speed of 20 km/h. How far can he travel in 4 hours?
Answer:
Mike can travel 80 Km in 4 hours
A fish swims to a depth of 50.00 meters in the ocean. Assuming the density of sea water is 1.0251.025 g·cm^{-3}g⋅cm −3 , calculate how much water pressure the fish is experiencing at this depth in units of kPa.
Answer:
The fish is experiencing a water pressure of 502.8 kPa.
Explanation:
The water pressure the fish is experiencing can be found as follows:
[tex]P = \rho gh[/tex] (1)
Where:
g: is the gravity = 9.81 m/s²
h: is the height (depth) = 50.0 m
ρ: is the seawater's density = 1.025 g/cm³
By replacing the above values into equation (1) we have:
[tex] P = \rho gh = 1.025 \frac{g}{cm^{3}}*\frac{1 kg}{1000 g}*\frac{(100cm)^{3}}{1 m^{3}}*9.81 m/s^{2}*50.0 m = 502.8 kPa [/tex]
Therefore, the fish is experiencing a water pressure of 502.8 kPa.
I hope it helps you!
Objects accelerate because
A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is the distance the bullet travels into the block?
How do you answer this question?
Answer:
d = 0.05 [m] = 50 [mm]
Explanation:
We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.
[tex]E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm][/tex]
Forces of 70 N at 130 degrees, and 20 N at an angle of 280 degrees, measured counter-clockwise from the positive x-axis, act on an object.
A. What are the components (F1x, F1y) of the first force force (in Newtons)?
B. What are the components (F2x, F2y) of the second force force (in Newtons)?
C. What are the components (Fx, Fy) of the resultant force (in Newtons)?
D. What is the magnitude of the resultant force (in Newtons)?
E. What is the angle of the resultant force with respect to x-axis?
Answer:
A. ) F₁ₓ = -45.0 N F₁y = 53.6 N
B.) F₂ₓ = 3.48 N F₂y = -19.7 N
C.) Fₓ = -41.5 N Fy = 33.9 N
D) F = 53.6 N
E) θ = -39. 2º (320.8º)
Explanation:
A)
Applying simple trig, like definitions of cos and sin of an angle, we can get the x- and y- components of F₁, as follows:[tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]
B)
Repeating for F₂:[tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]
C)
The x- and y- components of the resultant force, are just the algebraicsum of the x- and - y components of F₁ and F₂:
Fₓ = Fₓ₁ + Fₓ₂ = -45 N + 3.48 N = -41.5 N (5)By the same token, Fy can be written as follows:Fy = Fy₁ + Fy₂ = 53.6 N + (-19.7 N) = 33.9 N (6)D)
The magnitude of the resultant force can be obtained applying the Pythagorean Theorem to Fx and Fy, as follows:[tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]
E)
Finally the angle regarding the x- axis of the resultant force vector, can be obtained using the definition of the tangent of an angle, as follows:[tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]
What do light and energy tell us about the universe?
need a paragraph
How long must you wait (in half-lives) for a radioactive sample to drop to 2.10 % of its original activity?
Answer:
222/88 Ra
Explanation:
We have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.
To find the tike taken for the activity, we need to know about radioactivity and half-life.
What is radioactivity?Radioactivity is the rate of decay of a radioactive substance with respect to time. Mathematically, radioactivity is given asR=R₀e^(-λ×t)
From the above expression time is given ast= 1/λ ln(R₀/R)
What is half-life?Half-life is the time taken for decay of radioactive sample to half of its initial value. Mathematically, half-life= ln2 / λWhat is the expression of time of activity in term of half-life?From the half-life expression, 1/λ=half-life/ln2.Putting the value of 1/λ in the expression of time of activity, we havet=(half-life/ln2)×ln(R₀/R)
What is the time for radioactive sample to drop to 2.10 % of its original activity?Here R=0.021R₀, so t= (half-life/ln2)×ln(R₀/0.021R₀)=5.57 half-lives
Thus, we can conclude that we have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.
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Pushes and pulls that result from objects that are physically touching
each other
Answer:
That is false. Take a look at this way. You can push a ball with your own breath, you just need to blow it. And you can pull something from afar with a magnet. It is possible to do both.
Explanation:
Not all physical things can be done only physically. Like I just said, it is possible to use other forces (no, not the dark side one), such as a magnetic force, displayed by a magnet or anything with a force like so.
A monatomic ideal gas with an initial pressure of 500 kPa and an initial volume of 1.80 L expands isothermally to a final volume of 5.20 L. How much work is done on the gas in this process?
A) 1700J
B) 875J
C) 1570J
D) 900J
E) 955J
Answer:
955 J
Explanation:
PV = nRT
500 x 10³ x 1.8 x 10⁻³ = nRT
= 900 J
work done by gas in isothermal expansion
= nRT lnV₂ / V₁
= 900 ln 5.2 / 1.8
= 900 x ln 2.89
= 900 x 1.06
= 955 J
A diffraction grating with 68 slits per cm is used to measure the wavelengths emitted by hydrogen gas.
A. At what angles in the fourth-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm?
B. What are the angles if the grating has 12,800 slits per cm?
Answer:
a
[tex]\theta _1 =0.687 ^o[/tex]
[tex]\theta _2 =0.630 ^o[/tex]
b
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.1 will not be valid
Explanation:
From the question we are told that
The slit grating is [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]
The order of spectrum is [tex]n = 4[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 6800}[/tex]
=> [tex]a = 0.000147 \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _2 =0.630 ^o[/tex]
Gnerally if grating is [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 1280000}[/tex]
=> [tex]a = 7.813 *10^{-7} \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
[tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
One of the harmonics of a column of air in a tube that is open at both ends has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column in this tube?
Answer:
The fundamental frequency is [tex]f_1 =128 \ Hz[/tex]
Explanation:
From the question we are told that
The frequency of one harmonics is [tex]f_x= 448 \ Hz[/tex]
The next higher harmonic is [tex]f_z = 576 \ Hz[/tex]
Generally the frequency of an air column open at both ends is mathematically represented as
[tex]f_n = \frac{nv }{ 2 L }[/tex]
Here n is the order of the harmonics (frequency)
v is the velocity of the sound
L is the length of the column
So for one harmonics we have that
[tex]f_k = \frac{n v }{2L}[/tex]
Then for the next higher harmonics
[tex]f_x = \frac{n+1 ) v}{2 L }[/tex]
Generally the difference between these frequencies is mathematically represented as
[tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]
=> [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]
=> [tex]\frac{ v }{2L} = 128[/tex]
Generally for fundamental frequency n = 1
So
[tex]f_1 = n * \frac{v}{2L}[/tex]
So
[tex]f_1 =1 * 128[/tex]
=> [tex]f_1 =128 \ Hz[/tex]
Yellow light shines on a sheet of paper containing a blue pigment. Determine the appearance of the paper.
I dont need google answers if i get google answers i will delete it
Black I think
this is due to the fact that blue lights only reflect blue things so if yellow is shone on it it will reflect black appearance.
how it useful
Which of these statements is true about the International Space Station in orbit around the earth?
A) The space station exerts a force on the earth toward the space station.
B) The force of gravity produces the same acceleration for the earth and the space station. C) There is no gravity acting at the height of the space station.
D)The earth's gravity acts on the space station, but not the reverse.
Answer: b is probly correct beacuse all the others make no sence
The statement that is true about international space station is The force of gravity produces the same acceleration for the earth and the space station.
What is international space station?.International space station is a space station that is located at the lower part of the Earth's orbit . It is the third largest space object in the space and it is a multinational project that involves five different space agencies are NASA, Roscosmos, JAXA, ESA, and CSA.
Therefore, The statement that is true about international space station is The force of gravity produces the same acceleration for the earth and the space station.
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