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stepladder [879]
2 years ago
6

The resistance of a lamp that draws 0.5A of current when it is operated by a 12V battery will be Ω

Physics
2 answers:
Vedmedyk [2.9K]2 years ago
7 0
According to ohm's law
V=IR
Or R=V/I
R=12/0.5
R=24ohm
lisabon 2012 [21]2 years ago
5 0
According to Ohm's Law, the resistance, current and voltage are related as:

V = IR
⇒
R = V/I

V is given to be 12 Volts.
I is given to be 0.5 Ampere 

So, resistance will be:

R = 12/0.5 = 24 ohms
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Why are special techniques and extinguishing agents are required to fight combustible metals fires?
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3 years ago
The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s
just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c)  78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0
3 years ago
A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f
Butoxors [25]

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = \frac{p}{2l} \sqrt{\frac{T}{m} }

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 =\frac{3}{120} \sqrt{\frac{T}{2} }            I

Now if there is 5 antinodes , the value of p = 5

Thus n = \frac{5}{120} \sqrt{\frac{T}{2} }          II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

3 0
3 years ago
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