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2 years ago

A 12 v automobile battery is connected to an electric starter motor. the current through the motor is 246

1 answer:

6 0

The power dissipated by the motor is given by:
$P=VI$
where V is the voltage of the battery and I is the current flowing in the circuit.

In our problem, the voltage is V=12 V and the current is I=246 A, so we can calculate the power dissipated by the motor using the previous equation:
$P=(12 V)(246 A)=2952 W$

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Which is the most abundant element in the Sun?

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The answer is hydrogen.

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3 years ago

Vector A points along the + y axis and a magnitude of 100.0 units. Vector B points at an angle of 60.0 degrees above the +x axis

DedPeter [7]

Answer:

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2 years ago

A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s

Kryger [21]

The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

<u>Explanation:</u>

Given,

Temperature, T = 0°C

Initial mass, Mi = 62kg

Speed, s = 5.48m/s

Distance, x = 26.8m

Friction is present.

Mass of ice melted = ?

We know,

The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice

and

$Kinetic Energy, KE = \frac{1}{2} mv^2$

Therefore, $KE = \frac{62 X 5.48 X 5.48}{2}$

KE = 930.94 Joules

Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.

Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.

Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

4 0

2 years ago

The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter su

balandron [24]

The new period will be 2.486 days.

<h3>What is the period?</h3>

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

$\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days$

Hence, the new period will be 2.486 days.

To learn more about the period, refer to the link;

brainly.com/question/569003

#SPJ1

8 0

1 year ago

A 2.00 kg object is moving in a circular path with a radius of 5.00 cm. The object starts from rest and with constant angular ac

Rom4ik [11]

Answer:

0.800 m/s²

Explanation:

First, calculate the angular acceleration:

ω = αt + ω₀

6.00 rad/s = α (3.00 s) + 0 rad/s

α = 2.00 rad/s²

Now calculate the angular velocity at t = 2.00 s:

ω = αt + ω₀

ω = (2.00 rad/s²) (2.00 s) + 0 rad/s

ω = 4.00 rad/s

Calculate the linear velocity:

v = ωr

v = (4.00 rad/s) (0.0500 m)

v = 0.200 m/s

Finally, calculate the centripetal acceleration:

a = v² / r

a = (0.200 m/s)² / (0.0500 m)

a = 0.800 m/s²

3 0

3 years ago