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pogonyaev
2 years ago
8

A section of a rollercoaster is shown below. If the 17,500 kg rollercoaster momentarily comes to rest at point A, where it is 19

m in the air, how fast is it going at point D?
Physics
1 answer:
IgorLugansk [536]2 years ago
7 0

Answer:

Explanation:

Impossible to say without knowing how high point D is.

If we ignore friction, the energy converted from potential energy will exist as kinetic energy.

let d be the height in meters of point D "in the air"

mg(19 - d) = ½mv²

m is common so divides out

g(19 - d) = ½v²

v = √(2g(19 - d))

no sense in making g any more precise than the height. g = 9.8

v = √(2(9.8)(19 - d))

v = √(19.6(19 - d))

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1600 N
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5 0
3 years ago
An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (
lara31 [8.8K]

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV

Therefore, the average emf induced in the coil is 175 mV

3 0
3 years ago
What is the momentum of a baseball with a mass of 4 kg being thrown at a velocity of 84 m/s towards the hitter
Delvig [45]

Answer:

<h3>The answer is 336 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 4 kg

velocity = 84 m/s

We have

momentum = 4 × 84

We have the final answer as

<h3>336 kgm/s</h3>

Hope this helps you

7 0
3 years ago
Read 2 more answers
IF U ANSWER I WILL PUT U AS BRAINIEST ANSWER AND GIVE U A THANK U
Bingel [31]
<span>
At the Earth's surface, warm air expands and rises, creating
what is known as an area of low pressure.

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8 0
4 years ago
Read 2 more answers
A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B
VMariaS [17]

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

4 0
3 years ago
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