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hammer [34]
3 years ago
9

A car has a applied force of 120 N and experiences a friction force of 20 N. The mass is 20 kg. The acceleration is

Physics
2 answers:
GrogVix [38]3 years ago
4 0
Acceleration = 100/20 = 5 m/s^2
Alexandra [31]3 years ago
3 0
The answer is 5 m/s2
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5. Kelvin applied a rightward force of 302 N to a 29 kg desk to accelerate it across the floor. The
Naddik [55]

The force of friction is 213.2 N

Explanation:

The frictional force acting on an object sliding on a surface is given by:

F_f = \mu mg

where:

\mu is the coefficient of friction

m is the mass of the object

g is the acceleration of gravity

In this problem we have

\mu=0.750

m = 29 kg

g=9.8 m/s^2

Therefore, the force of friction is

F_f = (0.750)(29)(9.8)=213.2 N

Learn more about friction:

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3 years ago
Which is a chemical reaction? Select one: a. a peanut shell cracking open into two halves b. solid chocolate melting into liquid
raketka [301]
My best educated guess would be d 
8 0
3 years ago
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Ann [662]

Answer:

Answer:B

Explanation:

Because it all stayed consistant

4 0
3 years ago
A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accele
kobusy [5.1K]
The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:

v^2 = v_0^2 + 2ad\ \to\ a = \frac{-v_0^2}{2d} = -\frac{21.8^2\ m^2/s^2}{2\cdot 99\ m} = -2.4\frac{m}{s^2}

We can clear time in the speed equation:

v = v_0 + at\ \to\ t = \frac{-v_0}{a} = \frac{-21.8\ m/s}{-2.4\ m/s^2} = \bf 9.08\ s

If you find some mistake in my English, please tell me know.
3 0
3 years ago
Read 2 more answers
A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to
Svetach [21]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

         F = 5.20 \times 10^{5} N

         g = 9.8 m/s

         radius = \frac{diameter}{2}

                    = \frac{30 cm}{2} = 15 cm = 0.15 m   (as 1 m = 100 cm)

Formula to calculate depth is as follows.

        F = \rho \times g \times h \times A

or,      h = \frac{F}{\rho \times g \times A}        

       h = \frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}

           = 751 m

Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

3 0
3 years ago
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