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Law Incorporation [45]
3 years ago
5

What kind of friction exists between solid objects moving in water?

Physics
1 answer:
eduard3 years ago
6 0
Fluid Friction exists when it is acted upon an object when in fluid.


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HELP! A helicopter winches up an injured person from a boat. The person plus the equipment that must be pulled up with the perso
Marat540 [252]

When a person is pulled upwards why the motor of the helicopter

Then the potential energy increased  of the person is given as

U = mgh

here m = 109 kg

h = 83 m

so here potential energy of the person given as

U = 109*9.8*83

U = 88660.6 J

now power is defined as rate of energy

so it took 55 s to pull the mass so power is given as

P = \frac{U}{t} = \frac{88660.6}{55} = 1612 W

now we know that efficiency of motor is given as 69%

so we have

efficiency = \frac{output}{input}

0.69 = \frac{1612}{P}

so we have power as

P = \frac{1612}{0.69} = 2336.2 W


4 0
3 years ago
A riddle for you
koban [17]

Answer:

The Answer is FRICTION!

8 0
3 years ago
This question is about waves. I need help. If you answer can you explain why is that answer, thanks.
Dimas [21]

Answer:

b or d

Explanation:

i think ya sorry

4 0
3 years ago
when the armature of an ac generatr rotates at 15.0 rad/s, the amplitude of the induced emf is 27.0 V. What is the amplitude of
nata0808 [166]

To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

E = \omega NBA |sin \omega t|

Here,

\omega = Angular frequency

N = Number of turns

B = Magnetic field

The emf has its maximum value when sin \omega t = \pm 1

Thus the amplitude of the emf is

E = \omega NBA

When number of turns of armature, area and applied magnetic field remains constant, induced emf is proportional to angular speed.

E \propto \omega

Further it can be written as follows,

\frac{E_1}{E_2} \propto \frac{\omega_1}{\omega_2}

E_2 = \frac{\omega_2}{\omega_1}E_1

E_2 = \frac{10rad/s}{15rad/s}(27.0V)

E_2 = 18V

Therefore the maximum amplitude of induced emf when armature rotates at 10.0rad/s is 18V

5 0
3 years ago
A car of mass 940.0 kg accelerates away from an intersection on a horizontal road. When the car speed is 42.5 km/hr (11.8 m/s),
MAVERICK [17]

Answer:

0.39 m/s^2

Explanation:

Parameters given:

Mass of car, m = 940 kg

Speed of car, v = 11.8 m/s

Power supplied by engine, P = 4300 W

To get the acceleration, we must define the relationship between Power and velocity.

Power, P, is given in terms of velocity, v, as:

P = F * v

where F = force

This is because Power is given as:

P = \frac{E}{t} \\\\\\P = \frac{F*d}{t} \\\\\\=> P = F * v

(where E = energy. t = time taken, d = distance moved)

Force, F, is given as:

F = m*a

Therefore, Power will be:

P = m * a * v

Acceleration, a, will then be:

a = \frac{P}{m*v}

a = \frac{4300}{940 * 118} \\\\\\a = 0.39 m/s^2

The acceleration of the car at that time is 0.39 m/s^2

4 0
3 years ago
Read 2 more answers
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