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GrogVix [38]
3 years ago
5

A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels

Physics
1 answer:
spin [16.1K]3 years ago
7 0

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

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A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el
artcher [175]

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

                     p₀ = m v

The final moment

                     p_{f} = m (-v)

 

                  p₀ = - p_{f}

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

7 0
3 years ago
If you pull on something to the right with a force of 50 N and I pull on the same object to the left with a force of 60 N, what
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3 years ago
As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
Andreyy89

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

4 0
3 years ago
Read 2 more answers
A ramp is 1.0 m high and 3.0 m long. What is the IMA of the ramp?
oksano4ka [1.4K]
To calculate the ideal mechanical advantage for an inclined plane, divide th length of the incline by the height of the incline. 
Therefore; IMA = L/h
L= 3.0 m, while h =1.0 m
 IMA = 3/1
         = 3
Therefore the IMA of the ramp is 3 
This means the ramp increases the force that is being exerted by 3 times.
4 0
3 years ago
What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

7 0
3 years ago
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