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hichkok12 [17]
4 years ago
10

How much force is needed to stop a 80-kg football player if he decelerates at 5 m/s2

Physics
1 answer:
kiruha [24]4 years ago
6 0
According to Newton's 2nd law of motion, F =ma = 80x5 = 400N.
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A sample of a substance has a high density, yet a low particle motion. This sample must be a
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The part of the atom that accounts for electricity is the
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3 years ago
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Please guys i need help in this
JulijaS [17]

Answer:

10 seconds

Explanation:

We have the equation V = at  (speed = acceleration x time)

We want to find the time, so can rearrange to T = V/a (time = speed / acceleration).

From the question, we know V is 5 and a is 0.5.

Now we can substitute that into our equation: 5/0.5 = 10.

So the time is 10 seconds.

Hope this helps! Let me know if you have any questions :)

8 0
3 years ago
A pickup truck, initially stationary at a stop light, accelerates at a rate of 1.60m/s2 for 14.0 s. The truck then cruises at co
Alex Ar [27]

Answer:

The total distance traveled by the truck is 1797 m

Explanation:

Hi there!

The equation of position and velocity of an object moving in a straight line with constant acceleration are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the truck at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time

v = velocity at time t.

Let's calculate the position of the truck after the first 14.0 s:

x = x0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference at the first stop light, the initial position, x0, is zero. Since the truck starts from rest, v0 = 0. So, the equation of position will be:

x = 1/2 · a · t²

x = 1/2 · 1.60 m/s² · (14.0 s)²

x = 157 m

Then, the truck travels with constant speed (a = 0) for 70.0 s. The equation of position will be:

x = x0 + v · t

In this case, let's consider the initial position as the the position where the car is after 14.0 s (157 m from the stop light). The velocity is the velocity reached after the 14.0 s of acceleration. Let's calculate it with the equation of velocity:

v = v0 + a · t  (v0 = 0)

v = 1.60 m/s² · 14.0 s

v = 22.4 m/s

So, the position will be:

x = 157 m + 22.4 m/s · 70.0 s

x = 1725 m

Now, the truck slows down with an acceleration of 3.50 m/s² until it stops (until its velocity is zero). Let's calculate the time at which the velocity of the truck is zero:

v = v0 + a · t

0 = 22.4 m/s - 3.50 m/s² · t

-22.4 m/s / -3.50 m/s² = t

t = 6.4 s

Now let's calculate the position of the truck after that time considering the initial position as the position at which the truck was after the 70.0 s traveling at constant speed (1725 m from the stop light):

x = x0 + v0 · t + 1/2 · a · t²

x = 1725 m + 22.4 m/s · 6.4 s + 1/2 · (-3.50 m/s²) · (6.4 s)²

x = 1797 m

The total distance traveled by the truck is 1797 m

7 0
3 years ago
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