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blsea [12.9K]
3 years ago
7

4. A latch holding a cart in place, hanging high above the ground, breaks and the cart falls. The cart started from rest and is

in free fall until it hits the ground.
a. What is the velocity after 3.2 s?



b. How far has the cart fallen at 3.2 s?
Physics
1 answer:
scoray [572]3 years ago
7 0

Answer:

a) 31.4 m/s

b) 50.2 m

Explanation:

a) When an object is free falling, its speed is determined by the gravity force giving it acceleration. Equation for the velocity of free fall started from the rest is:

v = g • t

g - is gravitational acceleration which is 9.81 m/s^2, sometimes rounded to 10

t - is the time of free fall

So:

v = 9.81 m/s^2 • 3.2

v = 31.4 m/s ( if g is rounded to 10, then the velocity is 10 • 3.2 = 32 m/s)

b) To determine the distance crossed in free fall we use the equation:

s = v0 + gt^2/2

v0 - is the starting velocity (since object started fall from rest, its v0 is 0)

s = gt^2/2

s = 9.81 m/s^2 • 3.2^2 / 2

s = 50.2 m (if we round g to 10 then the distance is 10 • 3.2^2/2 = 51.2 meters)

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Answer:

A) Out of the page.

Explanation:

Right-hand rule points the direction of the magnetic field at any point.

<u>Top wire</u>: Current is to the left. Point your thumb to the left and curl your other fingers around the wire. The tips of the four fingers points the direction of the field at that point. In this case, out of the page.

<u>Bottom wire</u>: Current is to the right. Point your thumb to the right and curl your other fingers around the wire. The tips of the four finger points out of the page again.

So, the total field produced by both wires is directed out of the page.

Another method to figure out the direction is the mathematical method.

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The cross product between the direction of the current and the target position gives the direction of the B-field. If the left is -x direction and downwards is the -y direction, then

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3 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

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Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

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t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

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And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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