1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
erma4kov [3.2K]
3 years ago
12

A 6.20 g bullet moving at 929 m/s strikes a 850 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

in the same direction with its speed reduced to 478 m/s.
(a) What is the resulting speed of the block?
(b) What is the speed of the bullet-block center of mass?
Physics
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

(a) Final speed of block = 3.2896 m/s

(b) 6.7350 m/s is the speed of the bullet-block center of mass?

Explanation:

Given that:

Mass of bullet (m₁) = 6.20 g

Initial Speed of bullet (u₁) = 929 m/s

Final speed of bullet (v₁) = 478 m/s

Mass of wooden block (m₂) = 850g

Initial speed of block initial (u₂) = 0  m/s

Final speed of block (v₂) = ?

<u>By the law of conservation of momentum  as:</u>

<u>m₁×u₁ + m₂×u₂ = m₁×v₁ + m₂×v₂</u>

6.20×929 + 850×0 = 6.20×478 + 850×v₂

Solving for v₂, we get:

<u>v₂ = 3.2896 m/s</u>

Let the V be the speed of the bullet-block center of mass. So,

V = [m₁* u₁]/[m₁ + m₂]  (p before collision = p after collision)

   = [6.2 *929]/[5.2+850]

<u>V = 6.7350 m/s </u>

You might be interested in
You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle
AlekseyPX

Answer:

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

- mgh - F_f (s) = 0 - \frac{1}{2}mv^2

here we know that

F_f = \mu_k mg cos\theta

also we know that the length of the incline is given as

s = \frac{h}{sin\theta}

now we have

- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2

so we have

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

3 0
3 years ago
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/
Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0
3 years ago
Which of the following is numerically the same as the specific gravity? Mass Weight Density Volume
max2010maxim [7]
The answer is density

4 0
4 years ago
Read 2 more answers
A 53 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.36. The acceleration of gravity is 9.8 m
tino4ka555 [31]

Answer:

42.6 m

Explanation:

mass of crate m = 53 kg

coefficient of kinetic friction, μ = 0.36

acceleration due to gravity, g = 9.8 m/s^2

Force, F = 372.098 N

Net force, f = F - friction force

f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8

f = 185.114 N

acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2

initial velocity, u = 0

time, t = 4.94 s

s = ut + 1/2 at^2

s = 0 + 1/2 x 3.49 x 4.94 x 4.94

s = 42.6 m

6 0
3 years ago
Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express
AlladinOne [14]

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

          Vb = k Q / r        r <R

         

We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

         Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0
4 years ago
Other questions:
  • PLZ ANSWER ASAP
    13·1 answer
  • Please help on this one
    13·1 answer
  • This is my exam question be serious
    15·1 answer
  • Why is it a better to wear a white shirt on a hot summer day instead of a black shirt?
    13·2 answers
  • 4. A light string is attatched to a heavy rope, and the whole thing is pulled tight. A wave is sent along the light string. When
    7·1 answer
  • a train starts from rest and accelerates uniformly until it has traveled 2.1 km and acquired a forward velocity of 24 m/s.
    11·1 answer
  • Consider a car moving at a constant velocity. Recall that, for any object to move at a constant velocity, the net force acting o
    13·2 answers
  • 6. Compare Which of the
    6·1 answer
  • What are the symptoms of hepatitis 'b'​
    6·1 answer
  • Give three factors that affected the kinetic energy of the person as she reached the bottom
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!