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3 years ago

Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?

Physics

1 answer:

klasskru [66]3 years ago

6 0

Explanation:

It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.

We need to find the speed before and after the ball bounce.

Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

$\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s$

Let v is the final speed when it bounces to a height of 1.5 m. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s$

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.

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3 years ago

Read 2 more answers

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail

Alik [6]

Answer:

$1.34\cdot 10^{-16} C$

Explanation:

The strength of the electric field produced by a charge Q is given by

$E=k\frac{Q}{r^2}$

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

$E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C$

and the fish can detect the electric field at a distance of

$r=63.5 cm = 0.635 m$

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C$

4 0

2 years ago

Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m

ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

a = 62.5²/(2 × 22)

a = 88.78m/s²

the time you need to get this speed

v = v₀ + a t

t = v / a

t = 62.5 / 88.78

t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

t = 1,025 s

a = 55.43 m / s²

4 0

2 years ago