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
evablogger [386]
3 years ago
11

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

ion between the block and the plane is μk = 0.24. Determine the distance the block will slide before it stops.
Physics
2 answers:
nikklg [1K]3 years ago
3 0

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

Furkat [3]3 years ago
3 0

Answer:

The distance the block will slide before it stops is x= 3.4 m .

Explanation:

m1= 0,02 kg

V1= 400 m/s

m2= 2 kg

V2= ?

g= 9.8 m/s²

μ= 0.24

N = m2 * g = 19.6 N

Fr= μ * N

Fr= 4.704 N

due to the conservation of the amount of movement:

m1*V1 = m2*V2

V2= 4 m/s

Fr = m2*a

a= Fr/m2

a= -2.352 m/s²

matching momentum and amount of movement

Fr*t=m2*V2

t= 1.7 sec

x= V2*t - (a*t²)/2

x= 3.4 m

You might be interested in
Which are two types of mechanical energy electrical and kinetic nuclear and chemical radiant and thermal elastic and gravitation
WINSTONCH [101]
Elastic and gravitational.
5 0
3 years ago
Read 2 more answers
Why is an earthworm an invertebrate?
zimovet [89]

Earthworm have bones Some children confuse earthworms, which are invertebrates and have no bones, with snakes which are vertebrates and do have a bony skeleton and many ribs.

8 0
3 years ago
Read 2 more answers
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
(30 points) Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using v
Sophie [7]

Answer:

Explanation:

Check attachment for solution

8 0
3 years ago
A boulder released by a rockslide rolls 60 m in 4 s. What is its average speed?
Gelneren [198K]

Answer:

15 m per second

900m per minute

54,000 per hour

Explanation:

60 divided by 4 to get per second then times 60 for per minutes

then times 60 to get per hour

5 0
3 years ago
Other questions:
  • I need help with a physics test, can someone help i am willing to pay.
    11·1 answer
  • True or false : Scientific theories have withstood the test of time and are accepted as proven fact.
    11·2 answers
  • Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr
    15·1 answer
  • (HELP ASAP)
    10·2 answers
  • What does wap stand for? correct answer gets brainliest. NO LOOKING IT UP! THIS IS A TRIVIA!!
    9·2 answers
  • If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced
    9·1 answer
  • The net force acting on the ball below is ___ N
    9·1 answer
  • Which provides evidence of universe expansion that supports the Big Bang Theory? Select the three correct answers. (2 points)
    11·1 answer
  • At the bottom of a frictionless ramp, a 2kg mass reaches a speed of 9 m/s, how high is the top of the ramp? *
    6·1 answer
  • What is the equilibrium constant for the following reaction at 25 °c?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!