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Firlakuza [10]
3 years ago
13

A boat sails along the shore. To an observer, the boat appears to move at a speed of 11 m/s, and a man on the boat walking forwa

rd appears to have a speed of 12.4 m/s. How long does it take the man to move 6 m relative to the boat?
Physics
2 answers:
amid [387]3 years ago
4 0

You first subtract the speed at which the man is moving (11 m/s) from the rate the boat is moving (12.4 m/s). Which equals 1.4, then divide it by 6 meters, as the man is moving relative to the boat.

It therefore equals 4.29 s

zheka24 [161]3 years ago
4 0

Answer:

4.29 s

Explanation:

Apex

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A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving
Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

m_av_a + m_sv_s = 0

Where m_a = 92kg is the mass of the astronaut, m_s = 1200kg is the mass of the satellite, v_s = 0.14 m/s is the speed of the satellite. We can calculate the speed v_a of the astronaut:

v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0
3 years ago
The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

4 0
3 years ago
Read 2 more answers
Ceres, Pluto, and Eris are all round in shape and classified as:_________ A) Leftover planetesimals that formed inside the frost
motikmotik

Answer

Ceres, Pluto, and Eris are classified as DWARF PLANET.

A) Leftover planetesimals inside the frost line are known as ASTEROIDS.

B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.                                                

C) COMETS are the objects which are visible with long tails.            

D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.                

E) Meteor showers are associated with debris from COMETS                      

5 0
3 years ago
A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel
nexus9112 [7]

Answer:

The electric flux is 280\ \rm N.m^2/C

Explanation:

Given:

  • Radius of the disc R=0.50 m
  • Angle made by disk with the horizontal \theta=30^\circ
  • Magnitude of the electric Field E=713.0\ \rm N/C

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

\phi=\int E.dA

where

  • \phi is the total Electric Flux
  • E is the Electric Field
  • dA is the Area through which the electric flux is to be calculated.

Now according to question we have

=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C

Hence the electric flux is calculated.

8 0
3 years ago
Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On
8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

4 0
3 years ago
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