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yawa3891 [41]
4 years ago
12

Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend

ing around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth's magnetism (b) at points in space well above Earth's surface or (c) on Earth's surface? Round your answer (a) to 3 significant digits.

Physics
1 answer:
Lelechka [254]4 years ago
7 0

Answer:

Answer is

A. I = 6.3×10^8 A

B. Yes

C. No

Refer below.

Explanation:

Refer to the picture for brief explanation.

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The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre
IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0
3 years ago
An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work
Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ

Let W' is the work must be done to lift it 290 m. It is given by :

W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ

Hence, this is the required solution.

5 0
4 years ago
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in
qaws [65]

Answer:

Pressure increases due to enlargement

Explanation:

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

5 0
3 years ago
Read 2 more answers
A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat
jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

I=Ft

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

I=Ft=(40 N)(3.0 s)=120 Ns

8 0
3 years ago
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