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marshall27 [118]

3 years ago

12

What is the force of gravitational attraction between an object with a mass of 100 kg and another object that has a mass of 300

kg and are at a distance of 2m apart

1 answer:

Leokris [45]3 years ago

7 0

Answer:

5 x 10⁻⁷N

Explanation:

Given parameters:

Mass of object 1 = 100kg

Mass of object 2 = 300kg

Distance = 2m

Unknown:

Force of gravitational attraction between the objects = ?

Solution:

Newton's law of universal gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

From Newton's law of universal gravitation we derive an expression:

Fg = $\frac{Gm_{1} m_{2} }{r^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹

m is the mass

r is the distance between the bodies

Now insert the parameters and solve;

Fg = 6.67 x 10⁻¹¹ x $\frac{100 x 300}{2^{2} }$ = 5 x 10⁻⁷N

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What is the potential difference across a 15Ω resistor that has a current of 3.0 A?

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V = I • R

V = (15 ohms) x (3 A)

V = 45 volts

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

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(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

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3 years ago

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h

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Answer:

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Part b)

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Part e)

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Part f)

$\Delta K = 0$

Explanation:

As we know that initial velocity of the ball is given as

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Now final velocity of the system is given as

$v_f = 10.3\hat i - 5.72\hat j$

Part a)

now magnitude of initial momentum is given as

$P = mv$

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Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{2.93}{0.066}$

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Magnitude of change in momentum is given as

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Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

$\frac{\Delta P}{\Delta t} = F$

$\frac{5.02}{\Delta t} = 44.4$

$\Delta t = 0.113 s$

Part f)

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Which of these is an example of acceleration?

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Answer: A

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lorasvet [3.4K]

Answer:

the area of the rectangular field is 10.5 m²

Explanation:

Given;

length of the rectangular field, L = 42 cm = 0.42 m

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Area = 0.42 m x 25 m

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3 years ago