Question

solve this for 100 points work must be shown and must be all the problems PLZ I NEED IT ASAP​

Answer 1

1) Angular velocity: D) 73.3 rev/s

2) Centripetal acceleration: D) $5053 m/s^2$

3) Net force on the car: E) $2.19\cdot 10^4 N$

4) Orbital speed: D) $(\frac{GM}{r})^{\frac{1}{2}}$

5) Total mechanical energy: E) $-\frac{GMm}{2r}$

6) Gravitational force: C) $1.32\cdot 10^{-10}N$

7) Upward force: C) 188 N

8) Total upward force at B: C) 122 N

Explanation:

1)

The relationship between centripetal acceleration and angular velocity is given by

$a_c = \omega^2 r$

where:

$a_c$ is the centripetal acceleration

$\omega$ is the angular velocity

r is the radius of the circular path

In this problem,

$a_c = 1950 g = 1950(9.8)=19,110 m/s^2$

r = 9 cm = 0.09 m

Solving for $\omega$,

$\omega=\sqrt{\frac{a_c}{r}}=\sqrt{\frac{19,110}{0.09}}=460.8 rad/s$

And keeping in mind that $1 rev = 2 \pi rad$, we can convert into rev/s:

$\omega=460.8 rad/s \cdot \frac{1}{2\pi rad/rev}=73.3 rev/s$

2)

Again, the centripetal acceleration is given by

$a_c = \omega^2 r$

where in this problem we have:

$\omega=16 rev/s \cdot 2\pi (rad/rev) = 100.5 rad/s$ is the angular velocity of the piece of dust

r = 0.50 m is the distance of the piece of dust from the axis of rotation

And substituting into the equation, we find the centripetal acceleration of the piece of dust:

$a_c = (100.53)^2(0.5)=5053 m/s^2$

3)

Since the car is moving in circular motion, the net force is equal to the centripetal force, so it is given by:

$F=m\frac{v^2}{r}$

where

m = 732 kg is the mass of the car

v = 20.0 m/s is the linear speed of the car

r = 13.0 m is the radius of curvature of the loop-the-loop

Here we have

m = 732 kg

v = 20.0 m/s

r = 13.0 m

Substituting into the equation, we find:

$F=(732)\frac{20.0^2}{13.0}=2.25\cdot 10^4 N$

The closest option is E) $2.19\cdot 10^4 N$

4)

The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in circular motion, therefore we can write:

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where the term on the left is the centripetal force while the term on the right is the gravitational force, and where

m is the mass of the satellite

v is its speed

r is the orbital radius

G is the gravitational constant

M is the Earth's mass

Solving for v,

$v=\sqrt{\frac{GM}{r}} = (\frac{GM}{r})^{\frac{1}{2}}$

5)

The kinetic energy of the satellite is given by

$K=\frac{1}{2}mv^2$

And substituting its speed, $v=\sqrt{\frac{GM}{r}}$, we get

$K=\frac{GMm}{2r}$

The potential energy of the satellite is equal to the work that must be done against gravity to bring the satellite from r to infinite, and it is

$U=-\frac{GMm}{r}$

The total mechanical energy is equal to the sum of kinetic and potential energy:

$E=K+U=\frac{GMm}{2r}+(-\frac{GMm}{r})=-\frac{GMm}{2r}$

6)

The gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

Here we have:

$m_1 = 12 kg\\m_2 = 14 kg$

and

r = 9.22 m

Substituting,

$F=(6.67\cdot 10^{-11})\frac{(12)(14)}{(9.22)^2}=1.32\cdot 10^{-10}N$

7)

To solve the problem, we can apply the equation of the equilibrium of momenta about point B:

$Wx_W - F_A x_A=0$

where:

W = 300 N is the weight of the plank

$x_W = 5 m$ is the distance of the point of application of the weight from B (half the length of the plank)

$F_A$ is the upward force applied at A

$x_A = 10 - 2 = 8 m$ is the distance of the point of application of $F_A$ from B

Solving for $F_A$,

$F_A = \frac{W x_W}{x_A}=\frac{(300)(5)}{8}=188 N$

8)

We can solve this part instead by applying the equation of equilibrium of moments around point A:

$-Wx_W - W_B x_B + F_B x_{F_B} = 0$

where

W = 300 N is the weight of the plank

$x_W = 5 - 2 = 3 m$ is the distance of the point of application of the weight from A

$W_B = 9.0 N$ is the weight of the can of paint at B

$x_B = 10 - 2 = 8 m$ is the distance of the can of paint from A

$F_B$ is the upward force exerted by block B

$x_{F_B}= 10 -2 = 8 m$ is the distance of the point of application of $F_B$ from A

Solving for $F_B$,

$F_B = \frac{W_B x_B + W x_A}{x_{F_B}}=\frac{(9)(8)+(300)(3)}{8}=122 N$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

About gravitation:

brainly.com/question/1724648

brainly.com/question/12785992

About moments:

brainly.com/question/5352966

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