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GrogVix [38]
4 years ago
14

What do the length of the cord and gravity determine for a pendulum?

Physics
2 answers:
barxatty [35]4 years ago
5 0
Period is the correct answer
kap26 [50]4 years ago
5 0
The correct answer is period.
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The lightning bolt discharge lasts one millisecond. What is the average current flowing from the cloud to ground? Answer in unit
denpristay [2]

Answer:

25100A

Explanation:

t= 1ms = 0.001s

q = 25.1C

From the relationship between charge and current , the charge is equal to the product of current and time

q = i×t

Where q = charge

i = current

t = time

i = q/t = 25.1/0.001

i = 25100A.

5 0
3 years ago
Which of the following electrical components is a temporary electrical energy storage device?
7nadin3 [17]

Answer:

A capacitor

Explanation:

Because it can store electric energy when disconnected from its charging circuit. Commonly used in electronic devices to maintain power supply while batteries change.

Hope this helps! :)

6 0
3 years ago
A 10,000 N net force is accelerating a car at a rate of 5.5m/s^2. What is the cars mass
ycow [4]

Answer:

If a net horizontal force of 175 N is applied to a bike whose mass is 43 kg what acceleration is produced? What average net force is ... A 10,000 N net force is accelerating a car at a rate of 5.5 m/s2. What is the car's mass? A boy pedals his ...

Explanation:

6 0
3 years ago
With a 900 V source, voltage is divided across 3 series resistors of 300 V, 280 V, and:
telo118 [61]

Answer:

AAAAAAAAaaaaaaaaaaaaaaaaaaaaaaaaaaaa

6 0
3 years ago
An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

4 0
3 years ago
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