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-Dominant- [34]
4 years ago
7

In which situation is the object experiencing unbalanced forces?

Physics
1 answer:
s2008m [1.1K]4 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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Why is soil considered a non-renewable resource?
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Answer:D

Explanation: It is considered non renewable, it takes hundreds of years to renovate and because our life span as humans is very limited we don’t get to see how it regenerates.

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2 years ago
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A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the pro
svlad2 [7]

Answer:

Magnitude of the force on proton = F = 1.1085 × 10^-15 N

Explanation:

Charge on proton = q = 1.60 × 10^-19 C

Velocity of proton = V = 4.0 × 10^4 m/s

Magnetic field = B = 0.20 T  

Angle between V and B = θ = 60

We know that,  

F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)

F = 1.1085 × 10^-15 N    

8 0
3 years ago
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Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
3 years ago
If there was an unusually windy month, one might anticipate
Katarina [22]
Is there multiple answer choices?
8 0
3 years ago
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