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lora16 [44]
3 years ago
6

If you were trying to describe the difference between power and work you could say:

Physics
2 answers:
avanturin [10]3 years ago
6 0
Power is the energy transferred or "WORK DONE" in one second
nordsb [41]3 years ago
5 0

Answer:

Explanation:

Work done is defined as the product of force in the direction of displacement and the displacement.

Work = force in the direction of displacement x displacement

It is a scalar quantity and its SI uni is Joule.

Power is defined as the rate of doing work.

Power = Work / time

It is also a scalar quantity and its SI unit is Watt.

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As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax
AysviL [449]

Answer:

I=2.766\ kg.m^2

Explanation:

We have:

diameter of the wheel, d=0.88\ m

weight of the wheel, w_w=280\ N

mass of hanging object to the wheel, m_o=6.32\ kg

speed of the hanging mass after the descend, v_o=4\ m.s^{-1}

height of descend, h=2.5\ m

(a)

moment of inertia of wheel about its central axis:

I=\frac{1}{2} m.r^2

I=\frac{1}{2} \frac{w_w}{g}.r^2

I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

I=2.766\ kg.m^2

3 0
4 years ago
A real spring does not oscillate forever. Instead, it eventually comes to a stop. Does this violate the law of conservation of e
ss7ja [257]
It does not violate the law of conservation of energy. The oscillation stops when the energy is lost and the energy is lost because it becomes heat that is created by the air resistance and many other forces found in the surrounding of the oscillating spring.
5 0
3 years ago
What is Kepler’s first law?
BartSMP [9]

Answer:

Kepler's First Law: each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the orbital ellipse. The Sun is at one focus. The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit.

5 0
2 years ago
Read 2 more answers
A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o
Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j  D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of  the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the  projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0
3 years ago
A circular coil of 185 turns has a radius of 1.60 cm. (a) Calculate the current that results in a magnetic dipole moment of magn
s344n2d4d5 [400]

Answer:

(a) the current in the coil is 12.03 A

(b) the maximum magnitude of the torque is 0.0854 N.m

Explanation:

Given;

number of turns, N = 185

radius of the coil, r = 1.6 cm = 0.016 m

magnetic dipole moment, μ = 1.79 A·m²

Part (a) current in the coil

μ = NIA

Where;

I is the current in the coil

A is the of the coil = πr² = π(0.016)² = 0.000804 m²

I = μ / (NA)

I = 1.79 / (185 x 0.000804)

I = 1.79 / 0.14874

I = 12.03 A

Part (b) the maximum magnitude of the torque

τ = μB

Where;

τ is the maximum magnitude of the torque

B is the magnetic field strength = 47.7 mT

τ = 1.79 x 0.0477 = 0.0854 N.m

4 0
3 years ago
Read 2 more answers
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