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3 years ago

9

Last week we investigated the black hole at the center of our galaxy, with massMX= 8.5×1036kg. An object whose size is on the or

der of 1 m (it could beyou!) hangs at a distance from the center of the black hole. At what value of is the tidal force between one end of the object and the other greater thanthe gravitational force attracting the object to the black hole

1 answer:

swat323 years ago

6 0

Answer:

$d=2.38*10^{13}m$

Explanation:

We know the mass of the hole, so we define as,

$M_x= {8.5*10^{36}}$ kg

For which centripetal force is equal to gravitational force

$\frac{mv^2}{r}=\frac{GMm}{d^2}$

Angular velocity is equal to v/r,

$w^2r=\frac{6.67*10^{-11}*8.5*10^{36}}{d^2}$

As r=1 and w=1, we clear to d,

$d^2=5.6695*10^{26}$ m

$d=2.38*10^{13}m$

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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts

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(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

∑F is net force
m is mass of the third child
a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

the wagon
the children outside the wagon

<h3>Free body diagram</h3>

→ → Ф ←

1st child friction wagon 2nd child

<h3>Acceleration of the child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

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7 0

2 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

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3 years ago

A 1200 kg sports car accelerates from 0 m/s to 30 m/s in 10 s. What is the average power of the engine?

DIA [1.3K]

Answer:

3600N

Explanation:

Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s

ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

5 0

2 years ago

Read 2 more answers

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago