All categories

3 years ago

Explain briefly work energy and power

Physics

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

Work can be defined as transfer of energy.

Energy can be defined as the capacity for doing work

Power is the work done in a unit of time.

Explanation:

Work can be defined as transfer of energy. In physics we say that work is done on an object when you transfer energy to that object.

Power is the work done in a unit of time. In other words, power is a measure of how quickly work can be done.

Energy can be defined as the capacity for doing work. The simplest case of mechanical work is when an object is standing still and we force it to move

You might be interested in

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

Alisiya [41]

Answer:

The displacement of the same mass on the same spring on the Moon is 0.05 m.

Explanation:

Given;

mass suspended from one end of the spring, m = 0.500 kg

displacement on the spring on Earth, x = 0.3 m

Apply Newton's second law of motion;

F = ma = mg

where;

m is mass on the spring

g is acceleration due to gravity

Also, apply Hook's law;

F = Kx

where;

K is force constant

x is extension or diplacement of the spring

Combine the two equations from the two laws;

mg = kx

when the spring in on Earth;

0.5 x 9.8 = 0.3k

4.9 = 0.3k

k = 4.9 / 0.3

k = 16.333 N/m

when the spring is on moon;

mg = kx

mass is the same = 0.5 kg

acceleration due to gravity on moon = ¹/₆ that of Earth = ¹/₆ x 9.8 m/s²

0.5 (¹/₆ x 9.8) = 16.333 x

0.8167 = 16.333 x

x = 0.8167 / 16.333

x = 0.05 m

Therefore, the displacement of the same mass on the same spring on the Moon is 0.05 m.

4 0

3 years ago

Use Coulomb’s law to derive the dimension for the permittivity of free space. <br><br><br><br>

Paraphin [41]

Answer: M^-1 L^-3T^4A^2

Explanation:

From coloumb's law

K = q1q2 / (F × r^2)

Where;

q1, q2 = charges

k = constant (permittivity of free space)

r = distance

Charge (q) = current(A) × time(T) = TA

THEREFORE,

q1q2 = (TA) × (TA) = (TA)^2

Velocity = Distance(L) / time(T) = L/T

Acceleration = change in Velocity(L/T) / time (T)

Therefore, acceleration = LT^-2

Force(F) = Mass(M) × acceleration (LT^-2)

Force(F) = MLT^-2

Distance(r^2) = L^2

From ; K = q1q2 / (F × r^2)

K = (TA)^2 / (MLT^-2) (L^2)

K = T^2A^2M^-1L^-1T^2 L^-2

COLLEXTING LIKE TERMS

T^2+2 A^2 M^-1 L^-1-2

M^-1 L^-3T^4A^2

3 0

3 years ago

What happens to the ratio of surface area to volume as the cell size increases without the cell changing shape?

docker41 [41]

The volume increases, but the surface area remains the same

6 0

4 years ago

Read 2 more answers

What does the area under a curve of a graph of velocity versus time yield? A. Average acceleration O B. Average velocity OC.Tota

frutty [35]

Answer:

option (D)

Explanation:

If we plot a graph between the velocity of the object and the time taken, the slope of graph gives the value of acceleration of the object and the area under the graph gives the product of velocity and time taken that means it is displacement

5 0

3 years ago

Read 2 more answers

A jet plane is flying at a constant altitude. At time t1 = 0, it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t

Vika [28.1K]

Answer:

a) See attachment.

b) a_avg = -8.66 i - 2.33 j

c) a = 195 degrees from + x-axis. |a_avg| = 8.97 m/s^2

Explanation:

Given:

- Initial Velocity V_i = (90 i + 110 j)

- Final velocity V_f = (-170 i + 40 j)

- t_i = 0

- t_f = 30.0 s

Find:

(a) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?

(b) the components of the average acceleration

Solution:

- The change in velocity is given by:

Δ V = (V_f - V_i)

Δ V = ( -170 i - 90 i + 40 j - 110 j)

Δ V = (-260 i - 70 j)

- The change in time is given as:

Δ t = t_f - t_i = 30 - 0 = 30 s

-The average acceleration is:

a_avg = Δ V / Δ t

a_avg = (-260 i - 70 j) / 30

a_avg = -8.66 i - 2.33 j

- The magnitude of acceleration is:

|a_avg| = sqrt ( 8.66^2 + 2.33^2)

|a_avg| = sqrt (80.4245)

|a_avg| = 8.97 m/s^2

- The direction of acceleration is given by:

tan (Q) = a_y / a_x

Q = arctan(a_y / a_x)

Q = arctan(2.33 / 8.66)

Q = 15 degrees

Hence, 180 + 15 = a. a = 195 degrees from + x-axis.

8 0

3 years ago