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Advocard [28]
4 years ago
8

Checking Up 1. When a spring scale is used to do work pulling a cart to the top of an incline, where has the energy gone when th

e cart is at rest at the top? 2. Where does the roller coaster get its gravitational potential energy when it is at the top of the first hill? 3. Why do truckers use a ramp when loading a truck if the work required is the same with or without a ramp?
Physics
1 answer:
Kobotan [32]4 years ago
7 0

Answer:

1. When a spring scale is used to do work pulling a cart to the top of an incline, the elastic energy stored in the spring is converted into gravitational potential energy, since the top of an incline has a certain height.

2. The roller coaster get its gravitational potential energy from the kinetic energy it carries when it moves with a certain velocity.

3. Truckers use a ramp because it is easier to load a truck with less applied force. Although the work required is the same, without the ramp the force that they should apply is much greater, hence more difficult.

W = F_1 \times h\\W = F_2 \times d \times \cos(\theta)

h is greater than (d x cos(Θ)), hence F2 is smaller than F1.

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How did Niels Bohr change the atomic model based on his experimental results?
Alexus [3.1K]

Answer: Find the answer in the explanation

Explanation:

Bohr postulated that :

1.) Electrons revolve in specific circular orbits around the nucleus and these orbits can be represented by the letters K, L, M, N.

2.) Electrons revolving in a particular orbit are in a stationary state. They can neither gain energy nor loose energy. 3.) But if an electron jumps from an higher orbit to a lower orbit it will emit energy in the form of radiations. Also if an electron jumps from lower orbit to a higher orbit it excites and absorbs energy in the form of radiations.

4.) Neil Bohr experimental findings about angular momentum of an electron shows more light on the stability of atom and the line spectrum of hydrogen atom

8 0
4 years ago
A cyclist has a momentum of 771.9 kg m/s. if her mass is 93 kg what is the speed of the cyclist?
MrMuchimi
Take 771.9 Kg M/s  because that's how fast she is moving and divide it by 93 kg because that is her force pushing down

771.9/93= 8.3

Answer: A. 8.3 M/s

P.s I changed my answer on the last question you asked so change that.
3 0
4 years ago
Read 2 more answers
The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the eff
lara31 [8.8K]
We could use the change of pressure to calculate for the height climbed by the mountain hiker. The change of pressure is given by

p = rho * g * h, where p is the change of pressure, rho is the air density, g is the acceleration due to gravity, and h is the height.

Using the conversion 1 mbar = 100 Pa,

(930 - 780)(100) = (1.20)(9.80)h
15000 = 1.20*9.80*h

h = 1.28 km
6 0
4 years ago
A mountain climber increases their height from 200 meters to 400 meters. What affect will this have on their potential energy?
Yanka [14]

Answer:

At 400 m the potential energy of the mountain climber doubled the initial value.

Explanation:

Given;

initial height of the mountain climber = 200 m

final height of the mountain climber, = 400 m

The potential energy of the mountain climber is calculated as;

Potential energy, P.E = mgh

At 200 m, P.E₁ = mg x 200 = 200mg

At 400 m, P.E₂ = mg x 400 = 400mg

Then, at 400 m, P.E₂ = 2 x 200mg = 2 x P.E₁

Therefore, at 400 m the potential energy of the mountain climber doubled the initial value.

4 0
3 years ago
The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling frictio
levacccp [35]

Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

F_n = \rho A (v - u)^2 sin \theta

v is the velocity of jet

Q = A V

v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}

v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0

F n = 2295 N

now force along x direction

F_x = F_n sin 60^0

F_x = 2295 \times sin 60^0

F_x = 1987.52\ N

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

3 0
4 years ago
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