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Mariana [72]
3 years ago
15

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?
Physics
1 answer:
Svet_ta [14]3 years ago
6 0

Answer: a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

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state and explain any effect on sensitivity of liquid in a glass thermometer (i) reducing the diameter of capillary tube​
Ad libitum [116K]

Answer:

please give me brainlist and follow

Explanation:

The measuring sensitivity of liquid-in-glass thermometers increases with the amount of liquid in the thermometer. The more liquid there is, the more liquid will expand and rise in the glass tube. For this reason, liquid thermometers have a reservoir to increase the amount of liquid in the thermometer.

4 0
3 years ago
Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M
stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity M_1 g (downward) and the tension in the string T (upward). So we can write

M_1 g - T = M_1 a

where

M_1 is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, M_2 g sin \theta. So the equation of motion is

T-M_2 g sin \theta = M_2 a

where

M_2 = 13.5 kg is the mass of the 2nd block

\theta=35.5^{\circ} is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

a=0

So the two equations become:

M_1 g - T=0\\T-M_2 g sin \theta = 0

Isolating T from the 1st equation,

T=M_1 g

And substituting into the 2nd equation, we can find the value of the mass M_1:

M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg

Learn more about acceleration and forces:

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7 0
3 years ago
What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of?
madreJ [45]
We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) /  m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.

7 0
3 years ago
What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o
leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,  

d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}

This value can be used to determine the time for the pellet through the barrel:

t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s

Finally, we can use the above to calculate the force:

F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N



8 0
3 years ago
A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Figure 4-36, where t = 0
lana66690 [7]
In your question where as a golf ball is struck at a ground level and the speed of the ball as a function of time is in the figure where time t=0 and va = 16m/s and vb=32m/s. The following is the answer: 
a) How far does the golf ball travel horizontally before returning to ground level? 
-<span>80m</span>
<span>(b) What is the maximum height above ground level attained by the ball?
</span>-39.87m
3 0
3 years ago
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