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STALIN [3.7K]
3 years ago
11

A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The b

ullet embeds into the block. The speed, in m/s, of the bullet-plus-wood combination immediately after the collision is
Physics
1 answer:
ch4aika [34]3 years ago
6 0

Answer:

1.77 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............................ Equation 1

Where m = mass of the bullet, m' = mass of the block, u = initial velocity of the bullet, u' = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision

make V the subject of the equation

V = (mu+m'u')/(m+m').................. Equation 2

Given: m = 0.04 kg, m' = 6.96 kg, u = 310 m/s, u' = 0 m/s (stationary)

Substitute into equation 2

V = (0.04×310+6.96×0)/(0.04+6.96)

V = 12.4/7

V = 1.77 m/s

Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s

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Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
This picture represents the electric field diagram between two particles with static charges. Do the two particles have the same
dexar [7]

Answers:

No, They will attract each other, B, and neither direction

Explanation:

Since the two already presented particles in the diagram represent both opposing charges due to the direction of the arrows (the arrows facing away from the particle shows a positive charge and the particles facing towards the particle show a negative charge), not only because of this but as the arrows between the particles show an attracting magnetic field, then it can be concluded that the particles will attract to each other and if another particle was introduced into the diagram of a positive charge, then it would attract to the negatively charged particle. If you have any questions or need further explanation, please comment below. E2021, have a great day.

7 0
3 years ago
1) The smallest part of an element that behaves like the element is the atom.
Mandarinka [93]

Here are the answers: 

1. False - Molecules is the smallest part of an element that behaves like the element. 

2. False - The nucleus contains both protons and neutrons

3. True

4. True

5. A. Nucleus 

6. D. Neutron

7. B. Protons and Neutrons

8. C. Electron

9. C. 6

10. C.6

5 0
4 years ago
The person to introduce the idea of an atom was
Lelu [443]
C. Is the da answer to your question!
6 0
3 years ago
You are a crane operator using a wrecking ball to demolish an old building. You can choose to use a 100-kg ball or a 150-kg ball
Elenna [48]

Answer:

The answer to the question is

The two balls, although of different masses, could be made to have the same demolishing force by setting the velocity of the 100 kg ball to 1.5 times the velocity of the 150 kg ball.

That is if V₁ is the velocity of the 150 kg ball and  V₂ is the velocity of the 100 kg ball then V₂  = 1.5×V₁ for the demolishing effect of the two balls to be equal.

Explanation:

To answer the we are required to explain the meaning of momentum and state its properties

Momentum is a physical property of an object in motion. It indicates the amount of motion inherent in the object.  An object in motion is said to have momentum

The types of momentum possessed by an object can be classified into either

1, Linear momentum or

2. Angular momentum

An object moving with a velocity, v has linear momentum while a spinning object has an angular momentum

The momentum is given by the formula

P = m × V

Where m = mass and

V = velocity

Newtons second law of motion states that, the force acting on an object is equivalent to the rate of change of momentum produced and acting in the direction of the force

Properties of momentum

From the above statements it means that the two balls can be made equivalent by having the appropriate amount of speed. That iis the two balls can have the same momentum thus for equal momentum effect, we have

150 kg × V₁ = 100 kg × V₂

or V₂ = 1.5×V₁

3 0
3 years ago
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