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Alekssandra [29.7K]
2 years ago
11

A block of mass 5 kg slides down an inclined plane that has an angle of 10. If the inclined plane has no friction and the block

starts at a height of 0.8 m, how much kinetic energy does the block have when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2
Physics
1 answer:
Westkost [7]2 years ago
4 0

The kinetic energy of the block  when it reaches the bottom is 39.2 J.

<h3>Kinetic energy of the block at the bottom</h3>

Apply the principle of conservation of energy.

K.E(bottom) = P.E(top)

P.E(top) = mgh

where;

  • m is mass of the block
  • g is acceleration due to gravity
  • h is the vertical height of fall

P.E(top) = 5 x 9.8 x 0.8

P.E(top) = K.E(bottom) = 39.2 J

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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A certain x-ray tube requires a current of 7 mA at a voltage of 80 kV. The rate of energy dissipation is:
Artyom0805 [142]

Answer:

560 watts

The rate of energy dissipation is 560 W

Explanation:

Rate of energy dissipation is the rate of energy consumption in the x-ray tube.

P = VI

Given;

Voltage V = 80 kV = 80,000

Current I = 7mA = 0.007 A

Substituting the given values;

P = 80,000 V × 0.007A

P = 560 Joules per second

P = 560 watts

The rate of energy dissipation is 560 W

3 0
3 years ago
When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries?
Ket [755]

Answer:

Explanation:

Assume that a single baby cries at X db

4 babies crying would be X + 10log₁₀(4) = X + 6 db

8 0
2 years ago
According to the Second Law of Thermodynamics, a. any process during which the entropy of the universe increases will be product
Iteru [2.4K]

Answer:

The correct option is;

a. Any process in which the entropy of the universe increases will be product-favored

Explanation:

According to the second law of thermodynamics, the change in entropy of a closed system with time is always positive. That is the entropy of the entire universe, considered as an isolated system, always increases with time, hence the entropy change in the universe will always be positive.

\Delta S_{universe} = \Delta S_{system}  + \Delta S_{surroundings} >0

Therefore, any process in which the entropy of the universe increases will be product favored.

4 0
3 years ago
Frick and Frack are standing back-to-back, leaning on each other, but not moving. If Frick weighs
mylen [45]

Answer:

Frick is pushing harder

Explanation:

if Frack weighs more and he was pushing harder they would be moving, but if Frick pushes harder then they wont move

7 0
2 years ago
Read 2 more answers
Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.
balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

p_a=m_a.v_a

The total momentum of both pucks at the initial condition is

p_1=m_a.v_a+m_b.v_b

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

m_a=m

m_b=2m

We are given

v_a=6\ m/s\\v_b=2\ m/s

The total initial momentum is

p_1=6m+2(2m)=10m

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

p_2=3m.v'

Equating the initial and final momentum

10m=3m.v'

Solving for v'

v'=10/3\ m/s=3.33\ m/s

1. Compute the initial kinetic energy:

\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2

\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2

K_1=18m+4m=22m

The final kinetic energy is

\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2

\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2

K_2=16.63m

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

p_1=10m

p_2=3m.v'=3m(10/3)=10m

Proven

5 0
3 years ago
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