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andrey2020 [161]
3 years ago
15

500 J of work is used to decrease the angular velocity of a disk from 65 rad/s to 52 rad/s.What is the rotational inertia of the

disk?
Physics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

The correct answer is "0.66 kg.m²".

Explanation:

The given values are:

K.E = 500 J

w₁ = 65 rad/s

w₂ = 52 rad/s

As we know,

⇒  K.E=\frac{1}{2}I[w_1^2-w_2^2]

then,

⇒  I=\frac{2(K.E)}{w_1^2-w_2^2}

On putting the given values, we get

⇒     =\frac{2\times 500}{(65)^2-(52)^2}

⇒     =\frac{1000}{4225-2704}

⇒     =\frac{1000}{1521}

⇒     =0.66 \ kg.m^2

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Eva8 [605]
By using the telescope
4 0
3 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
A marble is placed at Location W. When the marble is released, it rolls down the track.
Norma-Jean [14]

Answer:

Z

Explanation:

6 0
2 years ago
Read 2 more answers
There is little vertical air movement in the __ because of the cold, heavy air at the bottom of the layer.
pogonyaev
The correct answer is the Mesosphere. <span>Mesosphere </span><span>is the layer of the Earth's atmosphere that is directly above the stratopause and directly below the mesopause.</span>
4 0
3 years ago
Read 2 more answers
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