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Digiron [165]
2 years ago
9

When a 100-N force acts horizontally on an 8.0-kg chair, the chair moves at a constant speed across the level floor. Which state

ment has to be true?
The chair accelerates at 12.5 m/s^2
The applied force is greater than the frictional force.
The friction force is 100 N
There is no friction between the floor and chair
Physics
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

The applied force is greater than the frictional force.

Explanation:

the chair moves at <u>a constant speed</u><u> </u><u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>not</u><u> </u><u>A</u><u> </u><u>or</u><u> </u><u>C</u><u>.</u>

if there is no friction then the chair <u>would accelerate and it would not be at a constant speed</u><u>.</u>

hence, the only possible answer is B.

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A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if
KengaRu [80]

Answer:

The value of  the correct angle of banking for the road is \theta = 67.76 °

Explanation:

Given data

Velocity (v) = 60 \frac{m}{s}

Radius = 150 m

The velocity of the car in this case is given by

v = \sqrt{r g \tan \theta}

v^{2} = r g \tan \theta

\tan \theta = \frac{v^{2} }{rg}

Put all the values in above formula we get

\tan \theta = \frac{60^{2} }{(150)(9.81)}

\tan \theta = 2.446

\theta = 67.76 °

Therefore the value of  the correct angle of banking for the road is \theta = 67.76 °

4 0
3 years ago
What is the weight of a 2.00-kilogram object on
Trava [24]
<span>(4) 19.6 N</span>
On Earth's surface, a mass of 1kg <span>exerts a force(weight) of 9.81 N
So the weight= 9.81x2.00
                      =19.62N
                      =19.6N</span>
3 0
3 years ago
Read 2 more answers
How many energy levels do the electrons occupy for a neutral Oxygen atom? *
Sophie [7]

Answer:

The neutral state of an atom is when it's net charge is zero; that is, the number of protons equals the numbers of electrons. Oxygen is the eighth element in the periodic table, with the symbol O. This means that it has eight electrons in its neutral state. Since it is neutral, it also has eight protons!

6 0
2 years ago
A car has a kinetic energy of 41.6 kJ.
musickatia [10]
K=1/2 mv2

M=?

41.6kj convert to joules by multiplying by 1000 so it will be 41,600J because the unit of kinetic energy is in joules.

41,600=1/2(m)(8)

Arrange the equation it will be:

M= 41,600/4 = 10,400

Final answer is:
m= 10,400 kg
4 0
2 years ago
A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e
Natali5045456 [20]

Answer:

a) F = 660.576\,N, b) a_{c} = 6.953\,\frac{m}{s^{2}}, c) v \approx 7255.423\,\frac{m}{s}, \omega = 9.583\times 10^{-4}\,\frac{rad}{s}, d) T \approx 1.821\,h

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

F = G\cdot \frac{m\cdot M}{r^{2}}

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}

F = 660.576\,N

b) The centripetal acceleration of the satellite is:

a_{c} = \frac{660.576\,N}{95\,kg}

a_{c} = 6.953\,\frac{m}{s^{2}}

c) The speed of the satellite is:

v = \sqrt{a_{c}\cdot R}

v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}

v \approx 7255.423\,\frac{m}{s}

Likewise, the angular speed is:

\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}

\omega = 9.583\times 10^{-4}\,\frac{rad}{s}

d) The period of the satellite's rotation around the Earth is:

T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)

T \approx 1.821\,h

6 0
3 years ago
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