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2 years ago

What is the relationship between Newton's second law and gravitational acceleration?

Physics

1 answer:

beks73 [17]2 years ago

5 0

Answer:

Correct answer: Fg = m · g

Explanation:

Newton's second law states that if a resultant force is applied to an object, the object begins to move at an accelerated rate.

The formula that presents this is:

F = m · a

this formula applies to an object moving on some surface

where m is the mass of the object and a the acceleration of the object

Let's take it now and watch the free fall:

The formula that presents this is:

Fg = m · g

this formula applies to an object moving at free fall in vertical direction

Free fall is also an accelerated movement to which Newton's second law applies.

where m is the mass of the object and g the gravitation acceleration of the object . We also know that g is equal:

g = γ · Me / d² where Me is mass of the earth

God is with you!!!

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Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what wil

drek231 [11]

Answer:The rate of ejection of photoelectrons will increase

Explanation:

If the frequency of incident monochromatic light is held constant and its intensity is increased, the rate of ejection of photoelectrons from the metal surface increases with increase in intensity of the monochromatic light. More current flows due to more ejection of photoelectrons.

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2 years ago

Playing soccer on a beach will require more effort because the sand causes a great deal of _______ to the ball.

34kurt

Answer:

I believe the answer to be B

Explanation:

If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry

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2 years ago

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A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha

jenyasd209 [6]

Answer:

Explanation:cuz i saiad so

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3 years ago

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

$v_y = 0$

While the horizontal velocity is

$v_x = 15 m/s$

The horizontal distance covered is

$d_x = 2.7 m$

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

$t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s$

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

$d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m$

8 0

3 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

3 years ago