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Natali [406]
2 years ago
14

A block of mass M is attached to a spring of negligible mass and can slide on a horizontal surface along the x-direction, as sho

wn above. There is friction present between the block and the surface. The spring exerts no force on the block when the center of the block is located at x=0. At the instant shown above, the block is located at x=0 and moving toward the right with speed v=v0. (a) For the instant shown above, with the block located at x=0 and moving to the right, predict the direction of the net force on the block. If the net force is zero, select “The net force is zero.”
_ To the left _ The net force is zero _ To the right
Briefly justify your prediction.
Physics
1 answer:
Hoochie [10]2 years ago
6 0

Answer:

Because of the frictional force, the net force will oppose direction of the block and be directed towards the left even tho the spring exerts no force at this point

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A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
3 years ago
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
Andy took a bus and then walked from his home to downtown. For the first 1.6 hour, the bus drove at an average speed of 15 km/h
Yuliya22 [10]

<em>1.6x15=24 </em>

<em> </em>

<em>0.4x4.5=1.8 </em>

<em> </em>

<em>24+1.8=25.8 </em>

<em> </em>

<em>1.6+0.4=2 </em>

<em> </em>

<em>25.8/2=12.9</em>

4 0
3 years ago
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A car travelling at 10 m/s accelerates at 3 m/s^2 for 6 seconds. What is the final velocity of the car?
tankabanditka [31]

Answer:28m/s

Explanation:I got one point so yeah hope this help

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3 years ago
I AM........ INEVITABLE
Archy [21]

Answer:

that's nice very nice super duper nicer

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