An engine is rated at 38% efficiency containing 415 J of energy in its heat reservoir. How much energy is in the

Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by

mylen [45]

Answer:

<em>I must travel with a speed of 2.97 x 10^8 m/s</em>

Explanation:

Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months

Time that elapses on the spacecraft = 1 year

On earth the people have advanced 120 yrs

According to relativity, the time contraction on the spacecraft is gotten from

$t$ = $t_{0} /\sqrt{1 - \beta ^{2} }$

where

$t$ is the time that elapses on the spacecraft = 120 years

$t_{0}$ = time here on Earth = 1 year

$\beta$ is the ratio v/c

where

v is the speed of the spacecraft = ?

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

120 = 1/ $\sqrt{1 - \beta ^{2} }$

squaring both sides of the equation, we have

14400 = 1/ $(1 - \beta ^{2} )$

14400 - 14400 $\beta ^{2}$ = 1

14400 - 1 = 14400 $\beta ^{2}$

14399 = 14400 $\beta ^{2}$

$\beta ^{2}$ = 14399/14400 = 0.99

$\beta = \sqrt{0.99}$ = 0.99

substitute β = v/c

v/c = 0.99

but c = 3 x 10^8 m/s

v = 0.99c = 0.99 x 3 x 10^8 = <em>2.97 x 10^8 m/s</em>

6 0

3 years ago

A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)

kondaur [170]

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s$

Initial pop up velocity is 35.87 m/s

a)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s$

It took 3.65 seconds to reach this height

6 0

3 years ago

A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

Ghella [55]

Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$

τ = $\dfrac{d}{dt}(10 - 3.5 t)$

τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

I = m R^2

I = 2 x 3.1²

I = 19.22 kg.m²

L = I ω

ω = $\dfrac{L}{I}$

ω = $\dfrac{10 - 3.5 t}{19.22}$

ω = $0.520 - 0.182 t$

A = 0.52 rad/s B = -0.182 rad/s²

5 0

3 years ago

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A string has its 4th harmonic at 31.5 Hz. What is the fundamental frequency?

seropon [69]

Given data

*The given 4th harmonic frequency is 31.5 Hz

The fundamental frequency is calculated as

$\begin{gathered} f_n=\frac{31.5}{4} \\ =7.875\text{ Hz} \end{gathered}$

Hence, the fundamental frequency is 7.875 Hz

5 0

1 year ago

A 3.1-kilogram gun initially at rest is free to

34kurt

The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
x=7.5/3.1
x=2.42

So the speed of the gun is 2.4m/s.

8 0

3 years ago

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