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natulia [17]
3 years ago
15

A box can slide with negligible friction on a horizontal table top. a string of negligible mass runs horizontally over a dowel w

ith negligible friction, and then connects to an identical box which is falling vertically. m m m~g how does the tension in the string between the boxes compare to the net force on the boxes? 1. it is less, about 50%.
Physics
1 answer:
Lina20 [59]3 years ago
3 0
<span>it is less, about 50%. Since we're ignoring friction, let's do the math and see what happens. I'll use the variables: M = mass of a single box. g = gravitational acceleration. The only force available is from the box that's suspended, so we'll have (1) Mg Newtons of force available. Now since the 2 boxes are connected together, that force will have to accelerate 2M. So the acceleration of the group is gM/2M = g/2 So both boxes are accelerating at half the local gravitational acceleration. This makes sense since the gravitational acceleration is only affecting half of the mass being accelerated. Now, to see the tension in the string, you need to ask the question, "How much force is required to accelerate the mass sliding horizontally at g/2?" The answer to that question is obviously Mg/2 newtons. So the net force being supplied is Mg newtons (See equation (1) above), and the force being transmitted via the string is Mg/2 Newtons, it's pretty obvious that the tension in the string is 50% of the total net force.</span>
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Answer:

Following are the answer to this question:

Explanation:

In option (a):

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In option (b):

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A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an
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Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say \omega

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

v=r\cdot \omega

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

(c)centripetal acceleration

Yes, there will be centripetal acceleration given by

a_c=\omega ^2\times r

                                   

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2 years ago
A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c
stealth61 [152]

Answer :

(a) The acceleration  of the car is, -2m/s^2

(b) The distance covered by the car is, 150 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = ?

Now put all the given values in the above equation 1, we get:

5m/s=25m/s+a\times (10s)

a=-2m/s^2

The acceleration  of the car is, -2m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = -2m/s^2

Now put all the given values in the above equation 2, we get:

s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2

By solving the term, we get:

s=150m

The distance covered by the car is, 150 m

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