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PIT_PIT [208]
3 years ago
6

9. To get the distance covered by an object, the speed by which it moves is

Physics
1 answer:
Liono4ka [1.6K]3 years ago
5 0

Answer:

time

................

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Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a
sergey [27]

Answer:

y = 104.4 ft

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

y = y_o + R + R sin(270 - 255)

y = y_o + R + R sin 15

here we know that

y_o = 10 ft

R = 75 ft

so we have

y = 10 + 75 + 75 sin15

y = 104.4 ft

4 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
I NEED HELP ASAP!!!
Savatey [412]
I think option C is correct..hope it helps
3 0
2 years ago
Read 2 more answers
An electromagnetic wave has a wavelength of 1.3m what is the frequency of wave
larisa86 [58]
V = f * wavelength
as we know electromagnetic wave has speed equal to light, so

3 * 10^8 = f * 1.3
f = 2.3 * 10^8 hertz
f = 230 mega hertz
6 0
3 years ago
A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
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