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kherson [118]
3 years ago
9

When a force of 15 newtons is applied to a stationary chair, it starts moving. What can you say about the frictional force betwe

en the chair and the floor of the room? A. The force of kinetic friction between the chair and the floor is greater than the force of static friction. B. The force of static friction between the chair and the floor is equal to the force of kinetic friction. C. The normal force acting on the chair is equal to 15 newtons in the upward direction. D. The force of static friction between the chair and the floor is less than 15 newtons.
Physics
1 answer:
Darina [25.2K]3 years ago
4 0

Answer:

D) the force of static friction between the chair and the floor is less than 15 newtons

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At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
An equipoterntial surface that surrounds a + 3.0 pC point charge has a radius of 2.0 cm. What is the potential of this surface?​
mestny [16]

Answer:

Electric potential = 0.00054 V

Explanation:

We are given;

Charge; q = 3 pC = 3 × 10^(-12) C

Radius; r = 2 cm = 0.02 m

Formula for the electric potential of this surface will be;

V = kqr

Where;

K is a constant = 9 × 10^(9) N⋅m²/C².

Thus;

V = 9 × 10^(9) × 3 × 10^(-12) × 0.02

V = 0.00054 V

8 0
3 years ago
Dr. james van allen and his group of united states scientists discovered a set of ___________. magnetic equators planets magneti
Fofino [41]

RADIATION BELTS....... I think but it should be radiation belt

3 0
3 years ago
Read 2 more answers
Does poop come out the pp or the butt please help im so confused best awnser gets brainleist
natulia [17]

Answer:

no poop comes out from your but

Explanation:

5 0
2 years ago
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horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
2 years ago
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