1

Superman takes 8 seconds to stop a runway train over a distance of 58 m using a power of

elena-14-01-66 [18.8K]

Answer:

Workdone = 600 Kilojoules

Explanation:

Given the following data:

Time = 8 seconds

Power = 75,000 Watts

Distance = 58 m

To find the work done;

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

Thus, work done is given by the formula;

Workdone = power * time

Workdone = 75000 * 8

Workdone = 600,000 = 600 KJ

5 0

3 years ago

Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, w

olga nikolaevna [1]

To solve this problem, apply the concepts related to the calculation of the work performed according to the temperature change (in an ideal Carnot cycle), for which you have to:

$W = \int\limit_{T_c}^{T_H} C (1-\frac{T_H}{T})$

Where,

C = Heat capacity of the Brick

$T_C$ = Cold Temperature

$T_H$ = Hot Temperature

Integrating,

$W = C (T_H-T_C)- T_H C ln (\frac{T_H}{T_C})$

Our values are given as

$T_H= 300K$

$T_C = 150K$

Replacing,

$W = (1) (300-150)-300(1)ln(2)$

$W = 150-300ln2$

$W = -57.94kJ \approx 58kJ$

Therefore the work perfomed by this ideal carnot engine is 58kJ

5 0

4 years ago

Plz solve this question..

ivann1987 [24]

Answer:

Explanation:

Voltage = 12 (the two batteries are in series and the voltage adds).

Resistance = 3 ohms

I = V / R

I = 12/3

I = 4 amperes.

None of the answers are the right ones. If there is a fifth one that reads 4 amperes then it is the correct answer.

4 0

3 years ago

A large cube has a mass of 25kg. It is being acclerated

soldier1979 [14.2K]

Answer:

p= 400.29N ........the horizontal force

Explanation:

Given data

mass=25 kg

small cube mass mass=4 kg

Us (The Coefficient of static b/w two cubes) = 0.71

to find

The horizontal force to keep the small cube from sliding downward

Solution

F=ma.........................from Newton Second law

Where F=force

a=acceleration

m=mass

we can write equation in form of acceleration

a=F/m

The acceleration on small box is same as that on the large box.

Let P be force to find.

then:

a=p/(25kg+4kg)

a=p/(29kg)m/s²

The force acting on small box:

F=ma

f=4*(p/29)N........................normal force

friction force= Us*(normal force).........where Us is coefficient of static friction.

friction force= 0.71*(4*p/29)

Now to find weight

weight= mg

weight= 4*9.8

for the object(small box) not to slide down the friction force b/w the two objects have to be exactly the same as the weight of the object.

0.71*(4*p/29)=4*9.8

solving for p(force)

p= 400.29N ........the horizontal force

3 0

3 years ago

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket

NARA [144]

The constant acceleration of a rocket launched upward, calculated knowing that the time it takes for a bolt that falls off the side of the rocker was 6.30 seconds, is 5.68 m/s².

When the rocket is launched straight up with constant acceleration, the acceleration of the rocket is given by:

$v_{f_{r}} = v_{i_{r}} + at$

Where:

$v_{f_{r}}$ : is the final velocity of the rocket

$v_{i_{r}}$ : is the initial velocity of the rocket = 0

a: is the acceleration

t: is the time

After 4 seconds, the final speed of the rocket will be the initial speed of the bolt, so:

$v_{f_{r}} = v_{i_{b}} = at = 4a$

When the bolt falls off the side of the rocket, the bolt hits the ground 6.30 seconds later.

The initial height of the bolt will be the final height of the rocket, and vice-versa. With this, we can take the final height of the bolt as zero.

$y_{f_{b}} = y_{i_{b}} + v_{i_{b}}t - \frac{1}{2}gt^{2}$

$0 = y_{i_{b}} + v_{i_{b}}t - \frac{1}{2}gt^{2}$

$y_{i_{b}} = \frac{1}{2}9.81*(6.30)^{2} - 4a*6.30 = 194.7 - 25.2a$

Now, as we said above, this height (of the bolt) will be the final height of the rocket, so:

$y_{f_{r}} = y_{i_{r}} + v_{i_{r}}t - \frac{1}{2}gt^{2}$

$194.7 - 25.2a = 0 + 0 - \frac{1}{2}a(4)^{2}$

$a = \frac{194.7}{33.2} = 5.86 m/s^{2}$

Therefore, the acceleration of the rocket is 5.68 m/s².

You can find another example of acceleration calculation here: brainly.com/question/24589208?referrer=searchResults

I hope it helps you!

3 0

3 years ago