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garik1379 [7]
4 years ago
8

A 2.0 kilogram rifle initially at rest fires a 0.002 kilogram bullet. As the bullet leaves the rifle with a velocity of 500 mete

rs per second , what is the momentum of the rifle -bullet system ?
Physics
1 answer:
Korolek [52]4 years ago
3 0

Answer:

1001

Explanation:

M=2.0+0.002=2.002kg

V=500m/s

p=?

p=M×V

p=2.002×500

p=1001

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A baseball is thrown straight up. the drag force is proportional to v2. part a in terms of g, what is the y-component of the bal
steposvetlana [31]

At the position of terminal speed the net acceleration of the ball will become zero

As we know that terminal speed will always reach when net force on the ball is zero and its speed will become constant.

So here at this position we can say

F_d = F_g

kv^2 = mg

v =\sqrt{\frac{mg}{k}}

now when ball is moving at half of the terminal speed in upward direction then net force on the ball in downwards direction will be

F_{net} = mg + F_d

F_{net} = mg + kv^2

here speed of the ball is half of the terminal speed

v = \frac{1}{2}*\sqrt{\frac{mg}{k}}

then we have

F_{net} = mg + k*\frac{mg}{4k}

F_{net} = \frac{5mg}{4}

now acceleration will be given as

a = \frac{F_{net}}{m}

now we have

a = \frac{5g}{4}

a = 12.25 m/s^2 downwards

7 0
4 years ago
Which of the following can be computed?
Musya8 [376]
Answer: only the third option. [Vector A] dot [vector B + vector C]

The dot between the vectors mean that the operation to perform is the "scalar product", alson known as "dot product".

This operation is only defined between two vectors, not one scalar and one vector.

When you perform, in the first option, the dot product of any ot the first and the second vectors you get a scalar, then you cannot make the dot product of this result with the third vector.

For the second option, when you perform the dot product of vectar B with vector C you get a scalar, then you cannot make the dot product ot this result with the vector A.

The third option indicates that you sum the vectors B and C, whose result is a vector and later you make the dot product of this resulting vector with the vector A. Operation valid.

The fourth option indicates the dot product of a scalar with the vector A, which we already explained that is not defined.
5 0
3 years ago
An object with charge -4.0μC is placed between a positively charged object to the
elena-s [515]

The net force on the -4.0μC object as a result of this net field will be -10.4 N.

<h3>What is the charge?</h3>

The matter has an electric charge when it is exposed to an electromagnetic field is known as a charge.

The electric field intensity is found as the force per unit charge.

Electric field intensity on the positive charge:

\rm E_1 = \frac{F_1}{Q} \\\\\ \rm   1.2 N/C = \frac{F_1}{ -4.0 \mu C} \\\\ F_1 =   - 4.8 \ N

Electric field intensity on the negative charge:

\rm E_2 = \frac{F_2}{Q} \\\\\ \rm   1.4 \ N/C = \frac{F_1}{ -4.0 \mu C} \\\\ F_2 =  - 5.6 \ N

The net charge is the algebraic sum of the two charges;

\rm F_{net}=F_1+F_2 \\\\ F_{net}= -4.8 -5.6 \\\\ F_{net}=-10.4 \ N

Hence, as a result of this net field, the net force on the -4.0C object will be -10.4 N.

To learn more about the charge refer to the link;

brainly.com/question/24391667

#SPJ1

4 0
2 years ago
A cell phone plan includes unlimited phone calls, texts, and data. There is an initial activation fee
Monica [59]

Answer:

90 months

Explanation:

1. 290 - 20 = 270

2. 270/3 = 90

4 0
3 years ago
Read 2 more answers
A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

    = 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0
3 years ago
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