Jaipal measures a circuit at 1.2 A and 24 . Using Ohm’s law, what can he calculate for the circuit?A. current (C)
B. current (I)
C. voltage (V)
D. voltage (A)

Answers

Answer 1

Answer:

Answer: The correct answer is Option C.

Explanation:

Ohm's Law is the law which has given a relationship between current, potential difference and the resistance in a circuit. This law states that the current flowing through the conductor is directly proportional to the potential difference across two points.

Mathematically,

$V\propto I$

or,

$V=IR$

where,

V is the potential difference across two points. It is expressed in Volts (V).

I is the current through the conductor. it is expressed in Amperes (A).

R is the resistance. It s expressed in Ohms $(\Omega)$

We are given:

$I=1.2A\nR=24\Omega$

By putting these values in above equation, we can easily calculate the value of voltage expressed in volts.

Hence, the correct answer is Option C.

Answer 2

Answer: The right answer for the question that is being asked and shown above is that: "C. voltage (V)" Jaipal measures a circuit at 1.2 A and 24 . Using Ohm’s law, he can calculate for the circuit is that through the C. voltage (V)

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A 50-kg ice skater is holding a 3-kg ball, moving at a steady +5 m/s. She then throws the ball backwardsat -10 m/s, if so what would her final velocity be?

Answers

Answer: 5.9 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_(o)$ must be equal to the final momentum $p_(f)$ :

$p_(o)=p_(f)$ (1)

Where:

$p_(o)=(m_(1)+m_(2))V_(o)$ (2)

$p_(f)=m_(1)U_(f)+m_(2)V_(f)$ (3)

$m_(1)=50 kg$ is the mass of the skater

$m_(2)=3 kg$ is the mass of the ball

$V_(o)=5 m/s$ is the initial velocity of the skater and the ball

$V_(f)=-10 m/s$ is the final velocity of the ball

$U_(f)$ is the final velocity of the skater

Substituting (2) and (3) in (1):

$(m_(1)+m_(2))V_(o)=m_(1)U_(f)+m_(2)V_(f)$ (4)

Isolating $U_(f)$ :

$U_(f)=((m_(1)+m_(2))V_(o)-m_(2)V_(f))/(m_(1))$ (5)

$U_(f)=\frac{(50 kg+3 kg)5 m/s}-(3 kg)(-10 m/s)}{50 kg}$ (6)

Finally:

$U_(f)=5.9 m/s$

The period of a ball moving in a circle is doubled while the radius stays the same. What happens to the velocity of the ball?

Answers

Answer:

The velocity of the ball becomes half when the time period is doubled.

Explanation:

Let $T$ be the time period of the ball, $R$ be radius and $v$ be the velocity of the ball performing circular motion.

Now, for a circular motion, the velocity is given as:

$v=(2\pi R)/(T)$

From the above relation, we can conclude that, velocity is inversely proportional to time period if the radius is constant.

This means that if time period is increased, velocity will decrease and vice-versa.

So, if time period is doubled, the velocity will reduce by half. This can be verified mathematically also given below.

$v_o=(2\pi R)/(T)\nv_n=(2\pi R)/(2T)$

Divide $v_n$ by $v_o$ , we get

$(v_n)/(v_o)=((2\pi R)/(2T) )/((2\pi R)/(T) )\n(v_n)/(v_o)=(T)/(2T)\n (v_n)/(v_o)=(1)/(2)\n\therefore v_n=(1)/(2)v_o$

Here, $v_n\ and\ v_o$ are the new and old velocities.

So, it can be concluded that if the time period is doubled, velocity is reduced by half.

How can a high school student have more momentum than a truck

Answers

Answer:

When truck is at rest while student is under motion

Explanation:

Since it is obvious that the mass of a truck is more than that of a student, we know that momentum is a product of mass and velocity

P=mv where m represent mass, v is velocity. When the student has more speed than that of truck, he exerts more momentum. The only way a student can exert more momentum is by having more speed while the truck is at rest. In such case, the momentum of truck will be zero while momentum of student will have a value

Does energy have mass and take up space?

Answers

Energy does not have a mass to take up space... because........let's say you shine a light into space the volume of the light beam is zero... And say you shine a torch by a black hole what would happen is gravity would pull the light due to gravity...so NO energy does not have have a mass and take up space....hope it helped

No its not true energy does have mass and take up space , that is matter so the real answer is matter

Provide 2 examples of viscosity

Answers

Low viscosity is fast-flowing, and the best example is water.

High viscosity is slow-flowing, like molasses or toothpaste.

Hot maple syrup has a LOWER viscosity than cold maple syrup!

viscosity is fast flowing.
high viscosity is slow flowing.
high viscosity example is toothpaste
low viscosity example is water

For the best answers, search on this site https://shorturl.im/awWwd

crude oil, for instance, shorter fractions have a lower viscosity than longer fractions (when separated in a condensing tower by fractional distillation)

Viscosity is a measure of the resistance of a fluid to being deformed by either shear stress or extensional stress. It is commonly perceived as "thickness", or resistance to flow. Viscosity describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. Thus, water is "thin", having a lower viscosity, while vegetable oil is "thick" having a higher viscosity. All real fluids (except superfluids) have some resistance to stress, but a fluid which has no resistance to shear stress is known as an ideal fluid or inviscid fluid. The study of viscosity is known as rheology.

Which phase change in the Atmosphere forms clouds?.

Answers

The phase is frequently used to describe transitions between the three fundamental states of matter. Condensation is the phase change in the atmosphere forms clouds.

What is phase transition?

The physical process of changing a medium's state from one to another is known as a phase transition in the domains of chemistry, thermodynamics, as well as other related sciences. The phase is frequently used to describe transitions between the fundamental states of matter.

The physical characteristics of a phase of such a thermodynamic system as well as the states of matter are constant. Some properties of a particular medium change throughout a phase transition as a result of a change in the environment, including such temperature or pressure. Condensation is the phase change in the atmosphere forms clouds.

Therefore, condensation is the phase change in the atmosphere forms clouds.

To learn more about phase transition, here:

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Condensation of the evaporated water

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