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Zielflug [23.3K]
2 years ago
8

A light bulb is plugged into a wall outlet (220 V) and it uses 0.68 amps. What is the light bulb’s resistance?

Physics
1 answer:
Veronika [31]2 years ago
5 0

Answer:

14.96 amps

Explanation:

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The noise floor, also known as additive white Gaussian noise (AWGN), is a continuous noise level that appears over a wide spectr
harkovskaia [24]

Answer:

correct option is a. True

Explanation:

solution

the noise floor is AWGN ( additive white Gaussian noise )  

and when viewed in the frequency domain, it is the continuous noise level  

because as they have a  uniform power over all the frequency.

 

so that it is additive white Gaussian noise  

as we can say given statement is True  

correct option a true  

4 0
3 years ago
A 50.0 kg person is walking horizontally with constant acceleration of 0.25 m/s² inside an elevator. The elevator is also accele
Alexxandr [17]

Answer:

The acceleration is in 2 D as in between east and south.

Explanation:

mass, m = 50 kg

acceleration, a = 0.25 m/s^2 horizontal

acceleration of elevator, a' = 1 m/s^2 downwards

When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:

6 0
2 years ago
A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?
Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

       A force of 55\text{ }N is acting on a wagon along the road. The wagon weights 7.5\text{ }kg. Acceleration of the wagon is given as 5.3\text{ }\frac{m}{s^{2}}.

       Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

       Gravitational Force and Normal Reaction cancel out each other.

       Net External Force = Mass of system/wagon \times Acceleration of wagon

       F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N

F_{friction} has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0
3 years ago
Please i need help! ill give brainliest toooo!
Igoryamba

Answer:

b

Explanation:

7 0
2 years ago
Read 2 more answers
Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict
GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = m

Mass of the heavier cart = 2m

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ KE =  \frac{mv^2}{2}

⇒ KE =  \frac{mv^2}{2}\times \frac{m}{m}

⇒ KE =  \frac{m^2v^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(mv)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ K=\frac{(F\times t)^2}{2m}           ...equation (i)

KE (K1) of mass 2m.

⇒ K_1=\frac{(F\times t)^2}{2\times 2m}

⇒ K_1=\frac{(F\times t)^2}{4m}         ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ \frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}

⇒ \frac{K_1}{K} =\frac{2m}{4m}

⇒ \frac{K_1}{K} =\frac{2}{4}

⇒ \frac{K_1}{K} =\frac{1}{2}

⇒ K_1=\frac{K}{2}

⇒ K_1=\frac{1}{2}K

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0
2 years ago
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