Which of the following is NOT an example of work?

Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object

vivado [14]

Answer:

C. The initial momentum should be equal to the final momentum due to the conservation of momentum.

$P_{initial} = mv_0\\P_{final} = (M+m)v_1\\v_1 = \frac{m}{M+m}v_0$

Since m/(M+m) < 1, v_1 > v_0.

Explanation:

Wrong -> A. Since the smaller particle still moves after the collision, it has a kinetic energy.

Wrong -> B. The total initial momentum is equal to the momentum of the smaller particle. Therefore, the momentum of the objects that stuck together is equal to that of the smaller object.

Wrong -> D. Since the bigger object is initially at rest and the surface is frictionless, the direction of motion will be the same as the direction of the smaller particle.

4 0

3 years ago

Why doesn't the principle of mechanical energy conservation hold in situations when frictional forces are present?

vodomira [7]

When you are talking about the Principle of mechanical Energy Conservation, it is really only including the kinetic and potential energy in a total system. When frictional forces are present, although the conservation of energy law is still present, it does not work when it comes to the conservation of mechanical energy as there is another type of energy that is factored in. As friction acts on the object, that transition from potential to kinetic as it slide/falls will be completely different as some of that energy is being transformed into thermal energy. Which breaks the conservation of mechanical energy.

7 0

3 years ago

A car starts at 12.5 m/s and reaches 25.5 m/s in 5.00 seconds. What is its acceleration? What is the car’s displacement? (Demons

disa [49]

The acceleration of a car that starts at 12.5 m/s and reaches 25.5 m/s in 5.00 seconds is 2.6m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a moving object can be calculated using the following formula:

a = (v - u)/t

Where;

a = acceleration (m/s²)
v = final velocity (m/s)
u = initial velocity (m/s)
t = time (s)

According to this question, a car starts at 12.5 m/s and reaches 25.5 m/s in 5.00 seconds.

a = (25.5 - 12.5)/5

a = 13.0/5

a = 2.6m/s²

Therefore, the acceleration of a car that starts at 12.5 m/s and reaches 25.5 m/s in 5.00 seconds is 2.6m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

4 0

2 years ago

HEY CAN YALL PLS ANSWER DIS!!!!!!!!

wlad13 [49]

Answer:

1.C

2.E

3.B

4.D

5.A

Explanation:

5 0

4 years ago

Read 2 more answers

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago