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3 years ago

If an object has 100 J of PE, how much work can it do (with that PE)?

Physics

1 answer:

Stella [2.4K]3 years ago

4 0

Answer:

W = 100 J

Explanation:

The potential energy of an object is given by :

$P=mgh$

It is said to possessed by an object due to its position under the action of gravity.

The work done by an object is given by :

W = F×d , F is force and d is displacement

We know that, F = mg

So, if an object has 100 J of PE, it means that it can do the work of 100 J.

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A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use

polet [3.4K]

The net force of the object is equal to the force applied minus the force of friction.
Fnet = ma = F - Ff
12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107.

7 0

3 years ago

Find the difference between two masses measured as 123.6 grams and 115.972 grams. Express the answer to the correct number of si

xxTIMURxx [149]

Answer:

The difference is 7.6 grams.

Explanation:

In mathematics the difference of two numbers is express as the subtraction between them:

$a-b$

So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.

Therefore, it is get:

$123.6grams-115.972grams = 7.6grams$

<u>Hence, the difference is 7.6 grams. </u>

The result of 7.628 will be expressed as 7.6 to have the correct number of significant figures.

Notice how that can be express in units of kilograms too since there is 1000 gram in 1 kilogram:

$7.6grams . \frac{1Kg}{1000grams}$ ⇒ $7.6x10^{-3}Kg$

8 0

3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

Radio waves are readily diffracted around buildings, whereas light waves are negligibly diffracted around buildings. This is bec

Tom [10]

Answer:

Radio waves have longer wavelength

Explanation:

Radio wave is an electromagnetic frequency that has the ability to travel through long distance. They have frequencies shuttling been the range of 10^4 hz and a frequency of 10^12 hz

Light wave is also called visible light. This is because it is visible to the naked eye, despite it being in the electromagnetic spectrum. It's frequency is usually between 4*10^-7 hz and a frequency of 7*10^-7 hz.

As can be seen from both, the radio waves length are quite far stronger than that of the light waves.

6 0

3 years ago

Noah stands 170 meters away from a steep canyon wall. He shouts and hears the echo of his

bezimeni [28]

340 ms

I got it right and I hope you do as well

6 0

3 years ago