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Arada [10]
3 years ago
8

In which scenario is the greatest amount of work done on a wagon?

Physics
2 answers:
ale4655 [162]3 years ago
5 0

Answer:

The first scenario!

Explanation:

W=F*d

a) 55*8= 440J

b) 60*6= 360J

c) 50*5= 250J

d) 40*10= 400J

Ray Of Light [21]3 years ago
4 0

Answer:

A. A force of 55 N moves it 8 m.

Explanation:

on edg

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What current is needed to generate the magnetic field strength of 5.0×10−5T at a point 1.5 cm from a long, straight wire? Expres
mixer [17]

Answer:

3.7 A

Explanation:

Parameters given:

Magnetic field strength, B = 5 * 10^(-5) T

Distance of magnetic field from wire, r = 1.5 cm = 0.015 m

The magnetic field, B, due to a current, I, flowing a wire is given as:

B = (μ₀*I) / 2πr

Where μ₀ = permeability of free space

To get the current, I, we make I the subject of the formula:

I = (2πr * B) / μ₀

I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))

I = 3.7 A

4 0
3 years ago
A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
A change in the gravitational force acting on an object will affect the object
MArishka [77]
It is weight, if I understand your question.
4 0
3 years ago
In which way are the cathode rays deflected in the x-plates of the Cathode Ray Oscilloscope​
MrRissso [65]

Answer:

The cathode ray is deflected vertically to the fluorescent screen

Explanation:

.

4 0
2 years ago
Read 2 more answers
The leg's force forward on the foot= 500N
Anna11 [10]

There's so much going on here, in a short period of time.

<u>Before the kick</u>, as the foot swings toward the ball . . .

-- The net force on the ball is zero.  That's why it just lays there and
does not accelerate in any direction.

-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.


<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...

-- The net force on the ball is 500N.  That's what makes it accelerate from
just laying there to taking off on a high arc.

-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward. 

That's how the leg's speed remains constant ... creating a dent in the ball
until the ball accelerates to match the speed of the foot, and then drawing
out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.


5 0
3 years ago
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