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Gnesinka [82]
3 years ago
13

A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn

itude of the carts acceleration if the time is 4.0?
Also if your really smart and have instagram please dm imaswagmeal and help me pls
Physics
1 answer:
olga55 [171]3 years ago
5 0

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

v_{f} = v_{i} + a*t\\

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

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A car moving at 60 mph slams on its brakes to stop before hitting a deer. Another identical car traveling at 60 mph slows to a s
ankoles [38]

Happy Holidays!

Recall that:

Impulse = Change in Momentum = mass × change in velocity

Since both cars are identical and have the same initial velocity of 60 mph, them breaking to a stop means that they both experience the same change in velocity.

Thus, both of the cars' impulses are equal.

8 0
3 years ago
What is the difference between speed and velocity?
avanturin [10]

Answer:

speed is the rate of change in distance thus it is scalar physical quantity

while velocity is the rate of change in displacement thus it is a vector physical quantity

Explanation:

vector physical quantity: is a quantity that requires both magnitude and direction to identify

scalar quantity: requires only magnitude to identify.

6 0
3 years ago
A train on a straight track goes in the positive direction for 5.9 km, and then backs up for 3.8 km
kenny6666 [7]

Answer:

2.1km

Explanation:

Ill take it as u are talking about the displacement

Since displacement has negatives and positves

5.9 - 3.8 = 2.1km

6 0
3 years ago
A transverse wave on a long horizontal rope with a
frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and  speed by the equation

f=\frac{v}{\lambda}

where

f is the frequency

v is the speed of the wave

\lambda is the wavelength

For the wave in this problem,

v = 2 m/s

\lambda=8 m

So the frequency is

f=\frac{2}{8}=0.25 Hz

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

T=\frac{1}{f}=\frac{1}{0.25}=4 s

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

t=\frac{T}{2}=\frac{4}{2}=2 s

4 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
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