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sveticcg [70]
3 years ago
12

Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,

how high could the pellet rise if fired on the Moon, where g
Physics
1 answer:
crimeas [40]3 years ago
4 0

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

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stich3 [128]
<h3><u>Answer</u> :</h3>

Initial velocity = zero (i.e., free fall)

Final velocity = 30m/s

Acceleration due to gravity = 10m/s²

For a body falling freely under the action of gravity, g is taken positive.

◈ <u>First equation of kinenatics</u> :

⇒ v = u + gt

⇒ 30 = 0 + 10t

⇒ t = 30/10

⇒ <u>t = 3s</u>

Hence, object will attain a speed of 30m/s after 3s.

8 0
3 years ago
A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .if after the collision it moves with a velocity of 2
ch4aika [34]

Answer:

-1.2 kg - m/s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .}\\\textsf{ After the collision it moves with a velocity of 2.0m/s in the opposite direction.}\end{cases}

And we need to find out the change in momentum of the body . Here ,

  • velocity before collision (u) = 10m/s
  • velocity after collision (v) = 2m/s .

We know that momentum is defined as amount of motion contained in a body . Mathematically ,

\sf\longrightarrow momentum (p)= mass(m) * velocity(v)

Therefore change in momentum will be,

\sf\longrightarrow \triangle p = mv - mu

Since the direction of velocity changes after the collision , the velocity will be -2m/s .

\sf\longrightarrow \Delta p = 100g( -2m/s -10m/s) \\

\sf\longrightarrow \Delta p =\dfrac{100}{1000}kg ( -12m/s)  \\

\sf\longrightarrow \Delta p   = 0.1 kg * -12m/s \\

\sf\longrightarrow \boxed{\bf \Delta p = -1.2 \ kg-m/s} \\

7 0
2 years ago
A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m
Alina [70]

Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

H = \frac{u^{2}Sin^{2}\theta }{2g}

2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

K=\frac{1}{2}mv^{2}

K = 0.5 x 3 x 3 x 3

K = 13.5 J

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Answer:

Water is liquid

Rock is solid

Thanks <3

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