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-Dominant- [34]
3 years ago
7

In which situation is the object experiencing unbalanced forces?

Physics
1 answer:
s2008m [1.1K]3 years ago
6 0

B. A car slowing as it reaches a stop light

Explanation:

According to Newton's second law, the net force exerted on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is its acceleration

This means that when an object is experiencing unbalanced forces (F\neq 0), it must have a non-zero acceleration. Therefore, let's analyze the different options:

A. A box resting on a horizontal floor  --> FALSE. Here the acceleration is zero, so no unbalanced forces.

B. A car slowing as it reaches a stop light  --> TRUE. Since the car is slowing down, its velocity is changing, so the acceleration is non-zero and there are unbalanced forces.

C. A car with its cruise control set to 50 mph  --> FALSE. Here the velocity is constant, so no acceleration and no unbalanced forces.

D. A rocket sitting on the launch pad --> FALSE. Here the rocket is stationary, so no acceleration, and no unbalanced forces.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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You are creating waves in a rope by shaking your hand back and forth. Without changing the distance your hand moves, you begin t
scoundrel [369]

Answer:

<u>Amplitude - remains the same</u>

<u>Frequency - increases</u>

<u>Period - decreases</u>

<u>Velocity - remains the same.</u>

<u />

Explanation:

The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.

The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.

The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.

The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.

7 0
3 years ago
A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates 1.5 m/s
Harrizon [31]
We now that follow newton rules f=ma so net force equal to mass*acceleration=>f=50*1.5=75 N
3 0
3 years ago
Read 2 more answers
Greg drew a diagram to compare two of the fundamental forces.
Mashcka [7]

Answer:

X: Always attractive

Y: Infinite range

Z: Attractive or repulsive

ANSWER IS C

8 0
3 years ago
Read 2 more answers
6) If a mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion
Zepler [3.9K]

Answer:

Decreases to half.

Explanation:

From the question given above, the following data were obtained:

Initial mass (m₁) = m

Initial force (F₁) = F

Initial acceleration (a₁) =?

Final mass (m₂) = ½m

Final force (F₂) = ¼F

Final acceleration (a₂) =?

Next, we shall determine a₁. This can be obtained as follow:

F₁ = m₁a₁

F = ma₁

Divide both side by m

a₁ = F / m

Next, we shall determine a₂.

F₂ = m₂a₂

¼F = ½ma₂

2F = 4ma₂

Divide both side by 4m

a₂ = 2F / 4m

a₂ = F / 2m

Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:

a₁ = F / m

a₂ = F / 2m

a₂ : a₁ = a₂ / a₁

a₂ / a₁ = F/2m ÷ F/m

a₂ / a₁ = F/2m × m/F

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

From the illustrations made above, the acceleration of the car will decrease to half the original acceleration

7 0
3 years ago
A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide
11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed  u = 2.00 m/s

Solution

Work = Change in kinetic energy

F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s

7 0
3 years ago
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