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Xelga [282]
3 years ago
7

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

. Assuming the acceleration is constant, how far did it travel during those 2.0 s?
Physics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

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Explanation:

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3 years ago
Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially
Gennadij [26K]

The trains take <u>57.4 s</u> to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

v_A=90 km/h\\ v_B=-80 km/h

The relative velocity of the train A with respect to B is given by,

v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

t= \frac{d}{v_A_B}

Substitute 2.71 km for d and 170 km/h for v_A_B

t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h

Express the time in seconds.

t=(0.01594h)(3600s/h)=57.39s

Thus, the trains cross each other in <u>57.4 s</u>.

6 0
3 years ago
Between ball B and ball C, ball ____ has LESS potential energy because ___
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Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat
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Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

8 0
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