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Gemiola [76]
2 years ago
15

This is a strong, short-lived magnet created by a looped electrical current.

Physics
1 answer:
Fed [463]2 years ago
4 0

Answer:elctromagnetic :D

Explanation:

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A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans
morpeh [17]

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

PE=mgh

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J

6 0
2 years ago
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What stores energy for a quick release in a cell?
luda_lava [24]
It is the mitochondria of a cell that stores energy for a quick release. <span>Mitochondria break down glucose to release the energy for cells to use. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
5 0
3 years ago
What kind of process can create "rogue" waves?
algol [13]
The correct answer for the question that is being presented above is this one: "a. Constructive interference." The kind of process that can create "rogue" waves is constructive interference. Constructive interference refers to <span>the </span>interference<span> of two or more waves of equal frequency and phase creating a mutual one.</span>
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2 years ago
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You can also describe position by using coordinates of _____and _______
SOVA2 [1]

Answer:

x and y

Explanation:

7 0
2 years ago
A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

6 0
3 years ago
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