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AysviL [449]
2 years ago
15

Drivers a and b travel west from boston. driver a leaves three hours earlier traveling at a constant speed of 68mph. driver b fo

llows at a constant speed of 85 mph. how many hours would it take for driver b to catch up to driver a?
Physics
1 answer:
Elza [17]2 years ago
5 0

We can answer this using one of the equations of linear motion:

v = d / t

where:

v = velocity

d = distance

t = time

<span>In the problem, we are asked to find for the time in which Driver B will catch up to Driver A. Therefore,  find the time when dA = dB. Rearranging the equation and equation dA and dB will result in:</span>

<span>vA * tA = vB * tB  ---> 1</span>

It was given that:

vA = 68 mph

tA = tB + 3 (since person A was travelling 3 hours earlier)

vB = 85 mph

tB = unknown

Substituting into equation 1:

68 * (tB + 3) = 85 * tB

68 tB + 204 = 85 tB

tB = 12 hrs

Therefore driver B would catch up to driver A after 12 hrs.

 

 

<span> </span>

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Explanation:

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A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o
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Answer:

Recoil speed, 1.17\times 10^{-5}\ m/s                          

Explanation:

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Mass of the comet fragment, m_1=1.96\times 10^{13}\ kg

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Mass of Callisto, m_2=1.08\times 10^{23}\ kg

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

m_1v_1=(m_2+m_2)V

V is recoil speed of Callisto immediately after the collision.

V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s

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7 0
3 years ago
which phrase best defines erosion. chemically breaking rocks apart, moving bits of rock from place to place , cementing bits of
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What are the parts of the lithosphere
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3 years ago
If two 1000kg cars have a gravitational force of 1.067x10^-5 N towards each other, how far apart are they?
Taya2010 [7]

Answer:

6.25m

Explanation:

Given parameters:

Masses of the car  = 1000kg

Gravitational force  = 1.067 x 10⁻⁵N

Unknown:

Distance between them  = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

     Fg  = \frac{G x mass 1 x mass 2}{distance^{2} }  

G = 6.67 x 10⁻¹¹

     1.067 x 10⁻⁵  = \frac{6.67 x 10^{-11} x 1000 x 1000 }{d^{2} }  

             d  = 6.25m

4 0
2 years ago
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