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2 years ago

The two ends of a coiled wire are connected to the electrodes of a lightbulb.

Physics

1 answer:

barxatty [35]2 years ago

5 0

Answer:

Explanation:

The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current.

in other words, the magnets motion creates a current in the loop

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1) The unit of power is

miss Akunina [59]

Answer:

1) the unit of power is A.Watt

2)100J=50W

100=50s

divide both sides by 50s

S=2s(A)

4 0

3 years ago

A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>

Mass of the box, m = 3.1 kg
Distance moved by the box, d = 2.5 m
Coefficient of friction, = 0.35
Inclination of the force, θ = 30⁰

Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

Learn more about work done here: brainly.com/question/8119756

8 0

2 years ago

A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric

MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

$J = \frac{I}{\pi R^2}$

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

$J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}$

This enclosed current is now to be used in Ampere’s Law.

$\mu_o I_{enc} = \int {B} \, dl$

Here, $\int \, dl$ represents the circular path of radius r. So we can replace the integral with the circumference of the path, $2\pi r$ .

As a result, the magnetic field is

$B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}$

5 0

3 years ago

What is the chemical name for lime

Sergio039 [100]

Calcium Oxide
CaO I think

5 0

3 years ago

A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.

Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s

Calculation of the acceleration of the hockey puck

We apply the Formula (1)

vf=v₀+a*t v₀=5.8 m/s , vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the frictional force (Ff)

We apply Newton's second law

∑F = m*a Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the hockey puck in the attached graphic

∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²

-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N = 1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0

3 years ago