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2 years ago

- What forces caused the rocket to no longer 'stay at rest' (its INERTIA)?

Physics

1 answer:

horsena [70]2 years ago

6 0

Answer:

Newtons first law

Explanation:

Objects stay at rest until a force acts on it causing it to move from its resting position

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A fan that is switched on for 1 minute uses 500 W usefully but also wastes 300 W through the emission of sound and heat. What's

koban [17]

Explanation:

Efficiency is defined as the ratio between the useful output over the total amount consumed. $E = \frac{useful}{total}$

The fan does 500W of useful work while wasting 300 W. The total power consumption is 800 W (500 + 300).

6 0

2 years ago

A 12.0 khz, 16.0 v source connected to an inductor produces a 4.00 a current. what is the inductance?

Varvara68 [4.7K]

Inductive reactance (Z) = ω L = 2Πf L = (2Π) (12,000) (L)

I = V / Z

4 A = 16v / (24,000Π L)

Multiply each side by (24,000 Π L):

96,000 Π L = 16v

Divide each side by (96,000 Π) :

L = 16 / 96,000Π = 5.305 x 10⁻⁵ Henry

L = 53.05 microHenry

4 0

2 years ago

A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e = v_c/m + vc_e

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

5m*vs_e - 4m*(v_c/m) = 5m*vc_e

vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

vc_e = 3760 - 0.8*(90)

vc_e = 3,688 km/h

5 0

3 years ago

Read 2 more answers

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$