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bogdanovich [222]
3 years ago
13

An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following pre

dictions is correct about why the object does not accelerate downward? Select two answers. Justify your selections.A) The bonded molecules of the object are repelled upward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.B)The normal force is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the objectC) The bonded molecules of the object are attracted downward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.D) The force of friction is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the object.
Physics
2 answers:
Montano1993 [528]3 years ago
6 0

Answer:

A) and B) are correct.

Explanation:

If the object is at rest, it means that no net force is exerted on it.

As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.

This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).

At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.

So, the statements A) and B) are true.

KIM [24]3 years ago
4 0

Answer:

The bonded molecules of the object are repelled upward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.

The normal force is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the object.

Explanation:

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An amplifier.

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Read 2 more answers
Calculate the power of an engine which can supply 60kg of energy per minute​
Nutka1998 [239]

Answer:

1 Watt

Explanation:

Power (P) = w / Δt

w = work done

Δt = change in time

Power (P) = 60kg / 60 seconds

Power (P) = 1 Watt

6 0
3 years ago
A resistor R and another resistor 2R are connected in a series across a battery. If heat is produced at a rate of 10W in R, then
alekssr [168]

Answer:

Option B

Solution:

As per the question:

Heat produced at the rate of 10 W  

The resistor R and 2R are in series.

Also, in series, same current, I' passes through each element in the circuit.

Therefore, current is constant in series.

Also,

Power,P' = I'^{2}R

When current, I' is constant, then

P' ∝ R

Thus

\frac{P'}{2R} = \frac{10}{R}

P' = 20 W

5 0
3 years ago
A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string,
Lorico [155]

Answer:

 v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

             y = y₀ + v_{oy} t - ½ g t²

             0 =y₀ + 0 -1/2 gt²

             t = \sqrt{ \frac{2y_o}{g} }

             t = √(2 40/32)

             t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

          sin θ = y / l

          θ = sin⁻¹1 y / l

          θ = sin⁻¹ 40/50

          θ = 53.1º

therefore the horizontal distance is

          x = l cos 53.1

          x = 50cos 53.1

          x = 30 m

let's use the equation

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 30 / 2.5

          v₀ₓ = 12 m / s

we look for the vertical component of the velocity

          v_y = v_{oy} - g t

          v_y = 0 - g t

          v_y = - 9.8 2.5

          v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

          v = \sqrt{v_x^2 + v_y^2}

          v = \sqrt{12^2 + 24.5^2}

          v = 27.28 m /s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹ 24.5 / 12

          θ = 63.9º

4 0
3 years ago
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