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MariettaO [177]
3 years ago
5

To maintain the same amount of torque due to a mass on a balance as the mass is increased, how should the position of the mass c

hange
Physics
1 answer:
olasank [31]3 years ago
8 0

Answer:

the mass should be bring closer to the point about which we are finding torque

Explanation:

τ = Σr × F = rmg

where m is the mass, g is acceleration due to gravity, and r is the distance

Torque is directly proportional to -

1.mass, m , of object

2. distance, r, of the mass from the point about which we are finding the torque.

So if we increase or decrease them then the torque will also increase or decrease.

So if we increase the mass the torque will increase but since we have to maintain same torque therefore we have to decrease the distance of mass from the point about which we are finding torque.

Therefore the mass should be bring closer to the point about which we are finding torque.

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Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring c
klemol [59]

Answer:

v_0=17.3m/s

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2

Since we only care about the velocity v_0, we can keep only the second and third parts of the equation and solve:

mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s

So, the speed of the ball just after the launch is 17.3m/s.

4 0
3 years ago
An object starts from rest at the origin and moves along the x axis with a constant
Alika [10]

Answer:

6.9m/s

Explanation:

Given parameters:

Acceleration of the object  = 4m/s²

Distance = from x; 2m to x; 8m

Unknown:

Average velocity  = ?

Solution:

From the given parameters, we use the right motion equation to solve the problem.

   v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

 S is the distance covered

 Distance  = 8m - 2m  = 6m

 Initial velocity  = 0m/s

The final velocity gives us the average velocity in this problem;

       v² = 0² + (2 x 4 x 6) = 48

      v = √48  = 6.9m/s

7 0
3 years ago
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Svetlanka [38]

Answer:

The dilation of time.

The falling of objects.

The changing of paths of light.

Explanation:

I have explained in the image attached below.

From the explanation, the correct ones are;

The dilation of time.

The falling of objects.

The changing of paths of light.

4 0
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Number order:

5, 4, 3, 2, 1

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it could lead to uncertainty of life and property

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