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Liono4ka [1.6K]
3 years ago
12

What is the net force acting on a golf car traveling at a constant speed of 5 mph

Physics
1 answer:
bekas [8.4K]3 years ago
4 0
Good homeworks
Good luck

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How does the mass of an object affect the outcome when an unbalanced force acts on it?
Verdich [7]

Answer:

The magnitude of acceleration is reduced.

Explanation:

Force is defined as push or pull

The force is said to be<em> balance force  </em>if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.

The force are to be<em> unbalanced force </em>if the force cause change in motion. ie the object has force greater than zero and has acceleration.

According to <em>Newton second law of motion </em>, acceleration depends on force acting on the object and mass of object.

     F=ma

     a=\frac{F}{m}

When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.

6 0
3 years ago
A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r
sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

         W =  Q_{Na^{+}}V

=>     W = 1.60 *10^{-19 } *  107 *10^{-3}    

=>     W = 1.712 *10^{-20 } \  J    

8 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb
balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = F_1=5.6 lb

Area of the piston = A_1=0.6 inches^2

Force applied on the brakes = F_2

Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

\frac{F_1}{A_2}=\frac{F_2}{A_2}

F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb

16.8 lb is the force on the brake pad of one wheel.

5 0
4 years ago
A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of
Sholpan [36]
F=ma
where:
F - force
m - mass
a - acceleration 

We transform this formula to get a:
a= \frac{F}{m}
a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
4 0
3 years ago
Read 2 more answers
Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the
xeze [42]
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is 
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
6 0
4 years ago
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