Which law relates to any pushing or pulling applied to the body to create motion?

Physics

2 answers:

STALIN [3.7K]3 years ago

5 0

A force relates to any pushing or pulling applied to the body to create motion.

bazaltina [42]3 years ago

3 0

Answer:

force.

Explanation:

I just took the test

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A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

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3 years ago

Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?

Schach [20]

Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.

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Which type of circuit is shown?

ivann1987 [24]

Answer:

I think it's D. open series circuit .

Explanation:

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