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2 years ago

12

An electromagnet is created using a battery, an insulated copper wire and an iron nail. The wire is wrapped around the nail 20 t

imes and attached to the battery. Which pair of changes will definitely increase the strength of the electromagnet?

1 answer:

Molodets [167]2 years ago

4 0

I think adding more loops would strengthen it

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What kind of star gives rise to a type i supernova?

larisa86 [58]

a. It occures in binary systems where one of them is a whitedwarf

3 0

3 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

2 years ago

"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a s

Tamiku [17]

Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

6 0

2 years ago

A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0

1 year ago

A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb

marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

$n_1 =\dfrac{46 \times 1}{0.0821\times 298}$

n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

$n_2 =\dfrac{12 \times 1}{0.0821\times 298}$

n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

$P_1=\dfrac{n_1 R T}{V}$

$P_1=\dfrac{1.89\times 0.0821\times 298}{5}$

P₁ = 9.25 atm

Partial pressure of oxygen

$P_2=\dfrac{n_2 R T}{V}$

$P_2=\dfrac{0.49\times 0.0821\times 298}{5}$

P₂ = 2.39 atm

total pressure

P = P₁ + P₂

P = 9.25 + 2.39

P = 11.64 atm

8 0

2 years ago