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3 years ago

Frequency refers to the number of wavelengths that pass a fixed point in a minute. true or false

Physics

1 answer:

Art [367]3 years ago

7 0

Frequency refers to the number of wavelengths that pass a fixed point in a minute. true or false

Answer:True

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A small, 3 kg weight is moved from 5 m from the ground to 8 m. What is the change in potential energy?

Ulleksa [173]

Answer: 88.2 J

Explanation: PE, OR Potential Energy, equals to mass gravity and height.

mass - 3kg

height - 3 because (8-5)

gravity is 9.8kg

so 3 times 3 times 9.8 = 88.2J

4 0

4 years ago

A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0

3 years ago

NEED HELP PLEASEEE 15 POINTS

Nezavi [6.7K]

Answer:

Explanation:

thamjs

7 0

3 years ago

Read 2 more answers

At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g

Strike441 [17]

Answer:

31.905 ft/s²

Explanation:

Given that

Mass of the pilot, m = 120 lb

Weight of the pilot, w = 119 lbf

Acceleration due to gravity, g = 32.05 ft/s²

Local acceleration of gravity of found by using the relation

Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²

3 0

3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

4 years ago