Which has the same meaning as air pressure?

An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

Afina-wow [57]

Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, $A=a^2=(0.305)^2=0.0930\ m^2$

Circumference of the loop,

$C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m$

Area of circle,

$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

The induced emf is given by :

$\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V$

So, the induced emf in the loop is $7.35\times 10^{-4}\ V$

8 0

3 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

A swimmer swims at 5 m/s. How long would it take to swim 5 laps of a 50m pool?

Gre4nikov [31]

About 5 hours gooood luck

3 0

1 year ago

State the five important assumptionns of the kinetic theory of matter

natulia [17]

ANSWER - (1) are constantly moving (2) have volume (3) have intermolecular forces (4) undergo perfectly elastic collisions (5) have an average kinetic energy proportional to the ideal gas’s absolute temperature

4 0

2 years ago

A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st

wel

Answer:

the tension T2 when the rock is completely immersed is T2 = 29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V= T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 = 29.05 N

3 0

3 years ago