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1 year ago

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Lamp power with what is the current through the lamp when it is connected to a battery ?

1 answer:

shtirl [24]1 year ago

4 0

Answer:

lamp power with what is the current through the lamp when it is connected to a battery ?

Explanation:

If the light bulbs are connected in parallel, the current flowing through the light bulbs combine to form the current flowing in the battery, while the voltage drop is 6.0 V across each bulb and they all glow. One bulb burning out in a series circuit breaks the circuit.

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A girl pushes a 60-kilogram box for a distance of 3 meters with a force of 4 Newtons into seconds. What is the power of the girl

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2 years ago

An air-filled capacitor stores a potential energy of 6.00 mJ due to its charge. It is accidentally filled with water in such a w

Pani-rosa [81]

This question is incomplete, the complete question is;

An air-filled capacitor stores a potential energy of 6.00 mJ due to its charge. It is accidentally filled with water in such a way as not to discharge its plates. How much energy does it continue to store after it is filled?

(The dielectric constant for water is 78 and for air it is 1.0006.)

Answer: it continue to store 0.07692 mJ after it was filled

Explanation:

Given that;

stored potential energy = 6.00 mJ = 0.006 J

dielectric constant for water K = 78

Energy stored U = Q² / 2C = 0.006 J

C = ∈₀A/d { Air}

C = K∈₀A/d { Water, k = 78 }

so

U = 0.006 / 78

U = 7.6723 × 10⁻⁵J

U = 0.07692 mJ

Therefore it continue to store 0.07692 mJ after it was filled

8 0

2 years ago

A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing

VARVARA [1.3K]

Explanation:

(a) The given data is as follows.

Length of the rod, L = 0.83 m

Mass of the rod, m = 110 g = 0.11 (as 1 kg = 1000 g)

At the lowest point, angular speed of the rod ( $\omega$ ) = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

I = $I_{CM} + mh^{2}$

= $\frac{1}{12}mL^{2} + m(\frac{L}{2})$

= $\frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}$

= 0.00631 + 0.415

= 0.42131 $kg m^{2}$

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

K = $\frac{1}{2}I \omega^{2}$

= $\frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}$

= 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b) According to the conservation of total mechanical energy of the rod, we have

$K_{i} + U_{i} = K_{f} + U_{f}$

$K_{i} = U_{f} - U_{i}$

or, mgh = K = 6.86 J

Therefore, h = $\frac{6.86}{mg}$

= $\frac{0.63}{0.11 \times 9.8}$

= 0.584 m

Hence, the center of mass rises 0.584 m far above that position.

6 0

2 years ago

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