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3 years ago

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]3 years ago

7 0

According to ohm's law
V=IR
Or R=V/I
R=12/0.5
R=24ohm

lisabon 2012 [21]3 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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A thunderclap sends a sound wave through the air and the ocean below. The

marysya [2.9K]

Answer:

C. 14.93 m

Explanation:

The given frequency of the wave, f = 100 Hz

The given equation for the wave speed, v, is presented as follows;

v = f × λ

The speed of sound in water, v = 1,493 m/s

Therefore, we get;

The wavelength, λ = v/f

∴ λ = 1,493 m/s/(100 Hz) = 14.93 m

The wavelength, λ = 14.93 m.

8 0

3 years ago

A rectangular object was found to have a mass of 1.278 kg and density of 4.98 g/cm3. Suppose that you knew that the length was

Leni [432]

Answer:

89.6 cm

Explanation:

From the question,

Volume of the rectangular object = Mass/Density.

V = m/D.................. Equation 1

Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³

Substitute into equation 1

V = 1.278/4980

V = 2.57×10⁻⁴ m³.

But,

V = lwh............... Equation 2

Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.

make h the subject of the equation

h = V/lw........... Equation 3

Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.

Substitute into equation 3

h = 2.57×10⁻⁴/(0.047×0.061)

h = 0.896 m

h = 89.6 cm

5 0

3 years ago

A car travels 30.0 km to the south and then turns around and travels 20.0 km to the north in 1.5 hours time. What is the car's t

mina [271]

Total displacement =10
Average velocity =6.7

5 0

3 years ago

Read 2 more answers

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

3 years ago

Avg velocity definition | Problem Average velocity vectors…

Aneli [31]

Answer:

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

Explanation:

As we know that average velocity is defined as the ratio of change in position and the time interval of the object

It is given as

$v_{avg} = \frac{\Delta x}{\Delta t}$

so we will have

$\Delta x$ = displacement

$\Delta t$ = time interval

the direction is this average velocity is always along the direction of the displacement

so correct answer will be

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

3 0

3 years ago