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Dafna1 [17]
3 years ago
13

When a loop of current-carrying wire turns continuously in a magnetic field, a _________ is created. A) galvanometer B) motor C)

transformer D) voltmeter
Physics
2 answers:
vlabodo [156]3 years ago
5 0
B) motor ... ;D I hoped I helped !
mixas84 [53]3 years ago
4 0

Answer: B motor.

As described in the question a loop of current carrying wire that is coil turns continuously that is rotations in a magnetic field. In a motor also there are rotations of current carrying loop in a magnetic field. There is no continuous rotations in galvanometer, transformer and Voltmeter only in motor there are continuous rotations.

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Which is heavier, 1 m3 of steel or 1 m3 of aluminium?​
Strike441 [17]

Answer:

Steel is almost 2.9 times heavier the aluminium.

5 0
2 years ago
Signment
jeka94

Answer:

v = 5.15 m/s

Explanation:

At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N

As the cable tension is less than this value, the car must be accelerating downward.

7730 = 984(9.81 - a)

a = 1.95 m/s²

kinematic equations s = ut + ½at² and v = u + at

-5.00 = u(4.00) + ½(-1.95)4.00²

u = 2.65 m/s    the car's initial velocity was upward at 2.65 m/s

v = 2.65 + (-1.95)(4.00)

v = -5.15 m/s

3 0
3 years ago
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2
Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂  = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0
2 years ago
A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran
marishachu [46]
<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train  = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train  =  4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

8 0
3 years ago
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