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Svetllana [295]

3 years ago

7

A 750-newton person stands in an elevator that isaccelerating downward. The upward force of theelevator floor on the person must

be(1) equal to 0 N (3) equal to 750 N(2) less than 750 N (4) greater than 750 N

1 answer:

Margaret [11]3 years ago

7 0

(2)<span>less than 750 N.( if the downward acceleration of elevator were g,then answer would be 0 N.)</span>

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2 years ago

A 7.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

Anvisha [2.4K]

Answer:

(a) 1.075 N

(b) 4.254 T

Explanation:

(a)

From the question,

Power = Force×Velocity.

P = Fv................... Equation 1

Where P = Power dissipated in the circuit, F = Pulling force of the wire, v = speed of the wire.

make F the subject of the equation

F = P/v................. Equation 2

Given: P = 4.30 W, v = 4.0 m/s.

Substitute into equation 2

F = 4.30/4

F = 1.075 N.

(b)

Applying,

F = BILsinФ.............. Equation 2

Where B = Strength of the magnetic field, I = current in the wire, L = Length of the wire, Ф = angle between the wire and the magnetic field.

make B the subject of the equation

B = F/ILsinФ................... Equation 3

But,

P = I²R...................... Equation 4

Where R = resistance of the wire.

make I the subject of the equation

I = √(P/R)............... Equation 5

Given: P = 4.30 W, R = 0.330 Ω

Substitute into equation 5

I = √(4.3/0.33)

I = √13.03

I = 3.61 A.

Also given: L = 7 cm = 0.07 m, Ф = 90°

Substitute into equation 3

B = 1.075/(0.07×3.61×sin90)

B = 1.075/0.2527

B = 4.254 T.

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3 years ago

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2 years ago

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a freight train travels at v=60(1-e^-t) ft/s, where t is the elapsed time in seconds. Determine the distance traeled in tree sec

lesantik [10]

Answer

given,

$v = 60(1-e^{-t})\ ft/s$

t = 3 s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx =\int 60(1-e^{-t}) dt$

$x = 60 t + 60 e^{-t}$

Position of the particle at t= 3 s

$x = 60\times 3+ 60 e^{-3}$

x = 182.98 ft

Distance traveled by the particle in 3 s is equal to 182.98 ft

now, particle’s acceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}60(1-e^{-t})$

$a = 60 e^{-t}$

at t= 3 s

$a = 60 e^{-3}$

a = 2.98 ft/s²

acceleration of the particle is equal to 2.98 ft/s²

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3 years ago

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3 years ago