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2 years ago

4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.

Physics

1 answer:

KIM [24]2 years ago

3 0

Answer:

The K.E of the bowling ball right before it hits the ground, K.E = 2450 J

Explanation:

Given data,

The mass of the bowling ball, m = 10 kg

The height of the building, h = 25 m

The total mechanical energy of the body is given by,

E = P.E + K.E

At height 'h' the P.E is maximum and the K.E is zero,

According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'

Therefore, P.E at 'h'

P.E = mgh

= 10 x 9.8 x 25

= 2450 J

Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J

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Which of the following describes the principle of conservation of charge?

grigory [225]

Answer: The statement "The charge cannot be created or destroyed describes the principle of the conservation of charge".

Explanation:

According to the conservation of charge, the charge can neither be created nor destroyed. It can be transferred from one system to another.

In an isolated system, the total electric charge remains constant. The net quantity of electric charge is always conserved in the universe.

Therefore, "the charge cannot be created or destroyed" describes the principle of the conservation of charge.

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3 years ago

What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? How far has the car traveled when it

nalin [4]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 60 mph = 26.8 m/s

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Use first equation of motion

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a = 2.68 m/s

Use second equation of motion

s = ut + 1/2 at²

s = 0 + 0.5 x 2.68 x 10 x 10

s = 134 m

As, 1 m = 3.28 ft

So, s = 134 x 3.28 ft

s = 439.6 ft

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3 years ago

How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?

Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

$d = \frac{ \:Δv}{e}$

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learn more about potential difference from here: brainly.com/question/28166044

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6 months ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

$f=\frac{33.84}{0.935}$

$f=36.19\approx 36\ \textup{Hz}$

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3 years ago

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