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2 years ago

Anyone know how this work ?? If u know help me it is an exam

Physics

1 answer:

myrzilka [38]2 years ago

6 0

Answer:

D) This is the correct answer

Explanation:

In this exercise the two ball loads are suspended by a thread.

To answer this exercise, let us remember that charges of the same sign repel and charges of a different sign attract.

Therefore, for the system to maintain equilibrium, the two charges must be of the same sign.

When examining the different proposals

A) in this case, as a sphere has no charge, there is no electric force and the induced charge is of the opposite sign, so the spheres attract each other

B) in this case there is an electric force, but being of a different sign, the force is attractive so the system is not in equilibrium

C) as the charges are of different magnitude the system does not have equal angles

D) This is the correct answer, since the charges have the same magnitude and are of the same sign, so the force is repulsive and is counteracted by the weight component

F_e = W sin θ

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .

Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

$v(1)= 6\,\,\frac{m}{s} \\$

5 seconds later the vehicle's velocity will be:

$v(5)=14\,\,\frac{m}{s}$

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration " $a$ "):

$v(t)=v_0+a\,t$

Therefore, in this case $v_0=4\,\,\frac{m}{s}$ and $a=2\,\,\frac{m}{s^2}$

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

$v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}$

3 0

2 years ago

Which of the following will be hardest to stop?

Dvinal [7]

Answer:OD. a hummingbird moving at 25 mph

Explanation:

3 0

3 years ago

Read 2 more answers

A man stands on his balcony, 130 feet above the ground. He looks at the ground, with his sight line forming an angle of 70° with

jenyasd209 [6]

Answer:

d = 380 feet

Explanation:

Height of man = perpendicular= 130 feet

Angle of depression = ∅ = 70 °

distance to bus stop from man = hypotenuse = d = 130 sec∅

As sec ∅ = 1 / cos∅

so d = 130 sec∅ or d = 130 / cos∅

d = 130 / cos(70°)

d = 380 feet

8 0

2 years ago

Describe how humans’ actions and natural processes have modified coastal regions in Louisiana and other locations

il63 [147K]

Coastal erosion has depleted a large portion of South Louisiana's wetlands along the coastline in swamps and marshes mainly due to storm surges. But other factors also contributed to this erosion. Canals and waterways dug through the marshes and swamps for the oil industry is one factor. Man-made levees erected to provide protection to residents living adjacent to the river is another major cause. Large scale logging especially in the early 1900's also damaged the wetlands.

3 0

3 years ago