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rosijanka [135]
3 years ago
6

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alys

sa starts 15 s later and runs at a constant 4.0 m/s. At what time after Jenny's start does Alyssa catch up with Jenny?
Physics
2 answers:
Delvig [45]3 years ago
8 0

Answer:

<h2>After 100 seconds Alyssa catch up with Jenny.</h2>

Explanation:

<h3>Jenny's data:</h3>

v_{Jenny} =3.8m/s

t_{Jenny}=t

d_{Jenny}=d

<h3>Alyssa's data:</h3>

v_{Alyssa}=4.0m/s

t_{Alyssa}=t-15, because she has a difference of 15 seconds.

d_{Alyssa}=d

Both move at a constant speed, that means there's no acceleration, their speed is always the same.

Now, the equation of each movement is

d=3.8t and d=4(t-15), then we solve this two.

We replace the first equation into the second one

3.8t=4(t-15)\\3.8t=4t-20\\20=4t-3.8t\\0.2t=20\\t=\frac{20}{0.2}\\ t=100

That means after 100 seconds Alyssa catch up with Jenny.

guapka [62]3 years ago
5 0

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s  t=57/0.2 =285 seconds

=4.75 minutes

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3 years ago
Two ice boats (one of mass m, one of mass 2m) hold a race on a frictionless, horizontal, frozen lake. Bath ice boats start at re
Elenna [48]

Answer:

They both cross the finish line with the same kinetic energy

Explanation:

Same force, same displacement, so same KE at the location of the finish line. They don’t cross the line at the same time, but that was not the question!

∆KA= (mA/2)vF,A2

∆KA=∆KBso vF,B/vF,A= (MA/MB)1/2∆KB= (mB/2)vF,B2

Lighter boat goes faster so reaches finish line 1st

5 0
3 years ago
How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?
saw5 [17]

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5 0
2 years ago
What is the radius of the Earth?
ki77a [65]
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7 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
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