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Strike441 [17]
2 years ago
14

An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of

the other what is the ratio of their masses
Physics
1 answer:
satela [25.4K]2 years ago
8 0
<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>
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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
2 years ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For maximum emf, \sin\omega t=1

So,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

So, the maximum emf generated in the loop is 5.32 V.

3 0
2 years ago
If 100 grams of vinegar and 5 grams of baking soda are poured in a container, a small amount of gas will be produced. What will
Anit [1.1K]
It will be a little bit less because of evaporation i learned that in third grade and your in high school that is sad
5 0
2 years ago
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
2 years ago
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