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noname [10]
3 years ago
13

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

ssa 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
1 answer:
MArishka [77]3 years ago
4 0

Answer:Answer:

300 seconds

Explanation:

We solve this problem using the equation of uniform motion;

d  =v*t

d: Distance traveled

t:  Time

v: Speed

We raise the equations of uniform motion because they run at a constant speed.

dJ=3.8*tJ   Motion equation for Jenny

dA=4*tA   Motion equation for Alyssa

d:distancia

When Jenny and Alissa meet, they will have traveled the same distance,then:

dJ=dA

3.8 tJ  =4*tA

tA =\frac{3.8}{4} * tJ

tA =0.95¨*tJ Equation (1)

Jenny runs 15 seconds longer than Alyssa,then:

tJ = tA+15 Equation (2)

We replace tA =0.95tJ of the Equation (1) in the Equation (2):

tJ = 0.95*tJ +15

tJ-0.95*tJ=15

0.05* tJ=15

tJ=\frac{15}{0.05}

tJ=300 s

Alyssa catch up with Jenny in 300 s after Jenny’s start .

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by energy of conservation

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A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is
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If the rock is completely submerged, the volume of the rock corresponds to the volume of water  displaced.

The volume of water displaced is given by

V_w = \frac{m_w}{\rho_w}

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\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3

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