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Ierofanga [76]
3 years ago
9

If you are measuring how far a ball traveled in an experiment which example below could be an

Physics
1 answer:
Galina-37 [17]3 years ago
3 0
20 meters long because the b
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A cruise ship is moving at constant speed through the water. The vacationers on the ship are eager to arrive at their next desti
MArishka [77]

Answer: 1. higher than it was before they started running

Explanation: As the vacationers run towards the back(stern) of the ship the exerting more pressure against the pressure exerted by the wave supporting the moving ship,the pressure exerted on the moving ship will be increased, leading to a slight increase in the speed of the ship compared to the speed before they started running towards the back(stern) of the ship.

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3 years ago
When energy from the sun hits the air above
valentina_108 [34]

Answer: C I think.

Explanation:

3 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
Please help me with this
Firlakuza [10]

Answer:

can't see anything sorry can't help

7 0
2 years ago
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While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba
Yuri [45]
Answer a would be correct since velocity is a vector and has a magnitude and a direction. In this case v₁ = - v₂.
8 0
3 years ago
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