Ask question

Ask question

All categories

3 years ago

8

An object is raised 2.5m off the ground. That object has a mass of 10kg, how many joules of potential energy is stored in the ob

ject? __ J

1 answer:

klio [65]3 years ago

4 0

Answer:

245.25 J

Explanation:

Potential Energy = m g h

= 10 * 9.81 * 2.5 = 245.25 J

You might be interested in

Two movers are pushing a large crate with a force of 60.0 n each. one pushes north, the other east. what is the equilibrant forc

romanna [79]

To get the solution you must need to draw a force triangle. Attach the head of the 60N north force arrow with the tail of the 60N east force arrow. The subsequent is the arrow connecting he tail and head of the two arrows.
You get a right angled triangle, and the resultant is (60^2 + 60^2) ^0.5 = 84.85 N or 85 N northeast.

8 0

3 years ago

The flux through the coils of a solenoid changes from 2.57.10-5 Wb to 9.44.10-5 Wb in 0.0154 s. If 4.08 V of EMF is generated, h

Vinil7 [7]

Hello!

We can use Faraday's Law of Electromagnetic Induction to solve.

$\epsilon = -N \frac{d\Phi_B}{dt}$

ε = Induced emf (4.08 V)
N = Number of loops (?)

$\Phi_B$ = Magnetic Flux (Wb)
t = time (s)

**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.

Now, we know that $\frac{d\Phi_B}{dt}$ is analogous to the change in magnetic flux over change in time, or $\frac{\Delta \Phi_B}{\Delta t}$ , so:
$\epsilon = N \frac{\Delta \Phi_B}{\Delta t}\\\\\epsilon = N \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}$

Rearrange the equation to solve for 'N'.

$N = \frac{\epsilon}{ \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}}$

Plug in the given values to solve.

$N = \frac{4.08}{ \frac{9.44*10^{-5} - 2.57*10^{-5}}{0.0154}} = 914.585 = \boxed{915 \text{ coils}}$

**Rounding up because we cannot have a part of a loop.

4 0

2 years ago

A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plan

gladu [14]

Answer:

(a) L = 0·73 m

(b) 4·39 × $10^{-3}$ J

Explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

<h3>For small oscillations, α = ω² × β</h3>

After substituting the value of α and solving we get

ω = √((3 × g) ÷ (2 × L))

Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))

∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4

Substituting the value of g as 9·8 m/s² and solving we get

L = 0·73 m

(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving

∴ Taking the reference for finding the potential energy as the lowest point

<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>

Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2

∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get

Maximum potential energy = 4·39 × $10^{-3}$ J

∴ Maximum kinetic energy = 4·39 × $10^{-3}$ J

4 0

3 years ago

GIVING BRAINLIEST !<br> answer quick !!!!!<br><br> What is the mass of a 1000 N person on Ea

REY [17]

Answer:

225-Ib person weighs 1000 N

7 0

3 years ago

to what part of the electromagnetic spectrum does the thermal radiation trapped in a greenhouse belong to?

IgorLugansk [536]

Answer:

Considerando, que al tope de la atmósfera llega un 100% de la radiación solar, de este total, sólo un 25% llega directamente a la superficie de la Tierra y un 25% es dispersado por la atmósfera como radiación difusa hacia la superficie, esto hace que cerca de un 50% de la radiación total incidente llegue a la ...

Explanation:

8 0

3 years ago