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3 years ago

What is the final speed if the displacement is increased by a factor of 4?

1 answer:

8 0

From the laws of motions:
x = 0.5 at^2 where
x is the displacement
a is the force of gravity (constant = 9.8 m/sec^2)
t is the time taken

Since "a" is constant, therefore:
the displacement is directly proportional to the square of the time.
This means that, increasing the displacement by a factor of 4 would increase the time by a factor of (4)^2 = 16.

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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

A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

$m_c$ = The center of mass of the stick = $\frac{L}{2}-0.22=0.5-0.22=0.28\ m$

$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

4 0

3 years ago

. What is the relationship between potential energy, kinetic energy, and speed as the skater moves down and up the U-shaped ramp

kykrilka [37]

Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest

Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest

THEREFORE, PE and KE have an INVERSE RELATIONSHIP.

6 0

3 years ago

Read 2 more answers

A $100 cart is moving down an inclined plane at 1.30 m/s. The cart collides with a stationary object and rebounds with a speed o

Rasek [7]

Answer:

0.65 kg*m/s and 0.165 kg*m/s

Explanation:

Step one:

given data

mass m= 0.5kg

initial velolcity u=1.3m/s

final velocity v= 0.97m/s

Required

The change in momentum

Step two:

We know that the expression for impulse is given as

Ft= mv

Ft= 0.5*1.3

Ft= 0.65 kg*m/s

The expression for the change in momentum is given as

P= mΔv

substitute

Pt= 0.5*(1.3-0.97)

Pt= 0.5*0.33

Pt=0.165 kg*m/s

7 0

2 years ago

Bx = -1.33 m and By = 2.81 m<br> Find the magnitude of the<br> vector.

NISA [10]

Answer:

Explanation:

The formula for the magnitude of a vector is

$B_{mag}=\sqrt{(-1.33)^2+(2.81)^2}$ and then round to the hundredths place:

3.11 m. Since we are in Q2, we can also find the direction of this vector:

$tan^{-1}(\frac{2.81}{-1.33})=-64.7$ but since we are in Q2, we add 180 degrees to the result, getting the angle to be 115.3

3 0

2 years ago