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WINSTONCH [101]
3 years ago
15

If everything lines up perfectly for superposition to occur, these massive waves would break with a force of 100 metric tons per

m". She compares this to how much force standing on a car?
Physics
1 answer:
Neko [114]3 years ago
4 0

Answer:

a

Explanation:

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How long did it take for a message sent from earth to reach the first astronauts?
Sergeu [11.5K]

Time taken for a message to sent from earth to reach the first astronauts is  140 second.

It takes about three days for a spacecraft to reach the Moon. in the course of that time a spacecraft travels at the least 240,000 miles 386,400 km that is the space between Earth and the Moon. The specific distance depends at the specific path selected.

The distance from earth to moon is approximately3.84*108m

So, the distance from earth to the astronauts is3.84*108m - 10500*103m =3.735*108m

The speed of light c = 2.99792458*108m/s

So, divided the distance from earth to astronauts by the speed off light,

3.735*108m/2.99792458*108m/s ˜1.25seconds

Divided the Pluto-earth distance by speed of light:

Hence, 4.2*1010m/2.99792458*108m/s = 140seconds

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8 0
1 year ago
A certain car's drive-train produces a force of 5300 N as it accelerates from 0
lakkis [162]

The power is 7.1\cdot 10^4 W

Explanation:

First of all, we need to find the acceleration of the car, which is given by

a=\frac{v-u}{t}

where

v = 60 mph = 26.8 m/s is the final velocity

u = 0 is the initial velocity

t = 10.0 s is the time

Substituting,

a=\frac{26.8-0}{10.0}=2.68 m/s^2

Now we can find the mass of the car by using Newton's second law:

F=ma

where

F = 5300 N is the force applied

m is the mass

a=2.68 m/s^2 is the acceleration

Solving for m,

m=\frac{F}{a}=\frac{5300}{2.68}=1978 kg

Now we can use the work-energy theorem, which states that the work done is equal to the change in kinetic energy of the car, to find the work:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2

And substituting,

W=\frac{1}{2}(1978)(26.8)^2-0=7.10\cdot 10^5 J

Finally, we can find the power output of the car:

P=\frac{W}{t}

where

W is the work

t = 10.0 s is the time elapsed

Substituting,

P=\frac{7.1\cdot 10^5}{10.0}=7.1\cdot 10^4 W

Learn more about power:

brainly.com/question/7956557

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4 0
4 years ago
A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact
Leno4ka [110]

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F =  average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

6 0
3 years ago
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic fie
Sliva [168]

Answer:

Please find the answer in the explanation

Explanation:

Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.

What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.

What happens above the coil?

the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines

Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.

Below the coil?

The needle will move in an opposite direction.

3 0
3 years ago
The kinetic friction coefficient between a cabinet and the floor is 0.3. Mass of the cabinet is 300kg. A man pushes the cabinet
podryga [215]

Answer:

<h2>0.39m/s^2</h2>

Explanation:

Step one:

given data

mass m= 300kg

applied force F= 1000N

coefficient of friction μ= 0.3

Step two:

The net force Fn= applied force-friction force  

Fn=F-F1

F1= limiting force

F1=μ*m*g

F1=0.3*300*9.81

F1=882.9N

the Net force= 1000-882.9

Fn=117.1N

Step three:

we know that

F=ma

Fnet=ma

a= Fnet/m

a=117.1/300

a=0.39m/s^2

7 0
3 years ago
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