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antiseptic1488 [7]

3 years ago

5

a student pushes a 40 in Block across the floor for a distance of 10 meters how much work was done to move the block A) 4j. B) 4

0j. C) 400j D) 4,000

1 answer:

Gemiola [76]3 years ago

4 0

Work = Force * distance
Work = 40N * 10m
Work = 400N

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alekssr [168]

Extensive lava flows

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3 years ago

Read 2 more answers

A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic

Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0

3 years ago

A certain pair of slits are separated by a distance d. Monochromatic coherent light falls on this pair of slits and the interfer

DanielleElmas [232]

Answer:

The new separation distance between adjacent bright fringes will be <u>4 mm</u>

Explanation:

Since, the distance between adjacent bright fringes is given by the formula:

Δx₁ = λL/d = 2 mm -------- eqn (1)

where,

Δx = Distance between adjacent bright fringes

λ = wavelength of light = constant for both cases

L = Distance between the slits and the screen

d = slit separation

Now, for the second case:

Slit Separation = d/2

Therefore,

Δx₂ = λL/(d/2)

Δx₂ = 2(λL/d)

using eqn (1), we get:

Δx₂ = 2 Δx₁

Δx₂ = 2(2 mm)

<u>Δx₂ = 4 mm</u>

5 0

3 years ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago

A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j

sladkih [1.3K]

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

Mechanical energy lost is equal top gain in kinetic energy

$ME_{Lost}=\dfrac{1}{2}mv^2$

$ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2$

$ME_{Lost}=662.29\ J$

b) Work done = Force x displacement

W = F. x

F = μ mg

W = μ mg . x

Work done is equal to 662.29 J

$x=\dfrac{W}{\mu m g}$

using the coefficient of the friction,μ = 0.7

$x=\dfrac{662.29}{ 0.7\times 70\times 9.8}$

x = 1.38 m

Hence, the runner will slide to 1.38 m.

3 0

3 years ago