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OleMash [197]
2 years ago
15

A toy remote control car drives in a circle (the controller is stuck). It makes 8 revolutions in 14

Physics
1 answer:
IceJOKER [234]2 years ago
7 0

Answer:

Period for 1 revolution is 1.75 seconds

Explanation:

given data

revolutions R = 8

time t = 14 seconds

to find out

What is the period

solution

we know that Period is the time per revolution

so here period formula that is express as

period = \frac{R}{t} = \frac{8}{14} = 0.57 revolution in one second

so in 1 revolution = \frac{1}{0.57} seconds

so in 1 revolution = 1.75 seconds

so period for 1 revolution is 1.75 seconds

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What mechanism of energy is transferred by mass motion of fluid from one region of space to another?​
lora16 [44]
Convection, because it is the process of heat transfer from one location to the next by the movement of fluids. The moving fluid carries energy within it.
5 0
3 years ago
A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co
makvit [3.9K]

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

5 0
2 years ago
Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k
Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

\frac{GM_{ns}}{R^2}= \omega^2 R

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0
3 years ago
An apparatus similar to the one used in lab uses an oscillating motor at one end to vibrate a long rope with frequency f = 40 Hz
Mice21 [21]

Answer:

The displacement in t = 0,  

y (0) = - 0.18 m

Explanation:

Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N

v = √ T / μ

v = √20.48 N / 0.02 kg /m = 32 m/s

λ = v / f

λ = 32 m/s / 40 Hz = 0.8

K = 2 π / λ

K = 2π / 0.8 = 7.854

φ = X * 360 / λ

φ = 0.5 * 360 / 0.8 = 225 °

Using the model of y' displacement

y (t) = A* sin ( w * t - φ )

When t = 0

y (0) = 0.25 m *sin ( w*(0) - 225 )

y (0) = 0.25 * -0.707

y (0) = - 0.18 m

5 0
3 years ago
Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
2 years ago
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