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Montano1993 [528]
3 years ago
7

A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat

es 10 m/sec2. What are the
combined masses of the two balls?
Physics
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

sorry I need points

Explanation:

IM SO SORRY

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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
3 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
2 years ago
Which of the following are true statements about the electric field? a. A moving charge produces a magnet anomaly. b. The electr
mars1129 [50]

Answer:

a,b,c,d,,f, g, j

Explanation:

e) equipotential lines are lines connecting points of equal potential

h) electric field inside the conductor is non-zero even when there is net movement of charge or non-zero current.

i) capacitors' plates are charged and an electric field exists between the plates.

4 0
2 years ago
How will creat thunderstrom​
emmasim [6.3K]

Answer:

the air has to be unstable as well as it needs to be moved upwards.

Explanation:

it needs to be moved upwards and also needs to have unstable air.

3 0
2 years ago
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