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White raven [17]
3 years ago
11

Your friend wants to be magician and intends to use Earth’s magnetic field to suspend a current-carrying wire above the stage. H

e asks you to estimate the minimum current needed to suspend the wire just above Earth’s surface at the equator (where Earth’s magnetic field is horizontal). Assume the wire has a mass of 10 g, and length of 1.0 m. Would you advise him to proceed with his plans for this act?
Physics
1 answer:
sesenic [268]3 years ago
7 0

Answer:

I=1960A

Explanation:

Under this  condition, . In order to suspend the wire, this magnetic force would  have to be equal in magnitude to the gravitational force exerted by Earth on the  wire the maximum force at angle

F=ILxB

Now to the suspend the wire so use the maximum force and solve to the current knowing the magnetic field of the earth

∑Fy=0

F_{m}-F_{g}=0

I*L*\beta-m*g=0

Solve to I current

I=\frac{m*g}{L*\beta}

I=\frac{10x10^{-3}kg*9.8m/s^2}{1m*0.5x10{-4}T}

I=1960A

I suggest do an toher act is really risk that current for an act

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Which of the following describes the measure of current
OleMash [197]
Um u need to show what the options are.
5 0
3 years ago
Read 2 more answers
A helicopter lifts a 60 kg astronaut 17 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/
dimulka [17.4K]

Answer:

a) W = 10995.6 J

b) W = - 9996 J

c) Kf = 999.6 J

d) v = 5.77 m/s

Explanation:

Given

m = 60 Kg

h = 17 m

a = g/10

g = 9.8 m/s²

a) We can apply Newton's 2nd Law as follows

∑Fy = m*a     ⇒     T - m*g = m*a     ⇒    T = (g + a)*m

where T is the force exerted by the cable

⇒    T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)

⇒    T = 646.8 N

then we use the equation

W = F*d = T*h = (646.8 N)*(17 m)

W = 10995.6 J

b) We use the formula

W = m*g*h     ⇒    W = (60 Kg)(9.8 m/s²)(-17 m)

⇒    W = - 9996 J

c) We have to obtain Wnet as follows

Wnet = W₁ + W₂ = 10995.6 J - 9996 J

⇒    Wnet = 999.6 J

then we apply the equation

Wnet = ΔK = Kf - Ki = Kf - 0 = Kf    

⇒  Kf = 999.6 J

d) Knowing that

K = 0.5*m*v²    ⇒    v = √(2*Kf / m)

⇒    v = √(2*999.6 J / 60 Kg)

⇒    v = 5.77 m/s

4 0
4 years ago
Read 2 more answers
A 10-kg projectile is fired straight up with an initial velocity of 500 m/s. (a) What is the projectile’s potential energy at th
Naily [24]
(a) the initial kinetic energy of the projectile is equal to:
K_i= \frac{1}{2}mv^2= \frac{1}{2}(10 kg)(500 m/s)^2=1.25 \cdot 10^6 J
The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to
U_f=K_i= 1.25 \cdot 10^6 J

b) If the projectile is fired with an angle of 45^{\circ}, its velocity has 2 components, one in the x-direction and one in the y-direction:
v_x = v_0 \cos 45^{\circ} =(500 m/s) \cos 45^{\circ} =353.6 m/s
v_y = v_0 \sin 45^{\circ} = (500 m/s)(\sin 45^{\circ} )=353.6 m/s

This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:
K_f =  \frac{1}{2}m v_x^2 = \frac{1}{2} (10kg)(353.6 m/s)^2=6.25 \cdot 10^5 J
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:
K_i = K_f + U_f
the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory:
U_f = K_i - K_f = 1.25 \cdot 10^6 J - 6.25 \cdot 10^5 J = 6.25 \cdot 10^5 J
7 0
3 years ago
When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it ta
bezimeni [28]

Answer:

Coefficient of static friction  ,μs = 0.15

Coefficient of kinetic friction  ,μk = 0.089

Explanation:

Given that

m = 1.7 kg

Static friction :

The force required to start the motion F= 2.6 N

We know that when book is in rest condition then static friction force act on it.

We know that the maximum value of the static friction force

fr = μ m g

At limiting condition

F= fr

2.6 =  μ m g

2.6 =  μ x 1.7 x 9.81

\mu=\dfrac{2.6}{1.7\times 9.81}

μ = 0.15

Kinetic friction :

F= 1.5 N

When the book is in moving condition then kinetic friction force act on it.

We know that the maximum value of the kinetic friction force

fr = μ m g

F= fr

=  μ m g

1.5 =  μ x 1.7 x 9.81

\mu=\dfrac{1.5}{1.7\times 9.81}

μ = 0.089

Coefficient of static friction  ,μs = 0.15

Coefficient of kinetic friction  ,μk = 0.089

7 0
3 years ago
Read 2 more answers
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

5 0
4 years ago
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