All categories

3 years ago

PLZZZ HELP!!! State the forces in a spring, magnets, electric scales, and a newton metre

Physics

1 answer:

Lisa [10]3 years ago

5 0

Answer:

The various kinds of forces that exist in the items below are:

Spring = Spring Force

Magnet = Magnetic Force

Electric Scale = Electric Force

Newton Metre = Spring Force

Explanation:

Spring Force:

When you fix an object unto or upon a spring, the resulting compression or stretching exerted by that object on the spring is called the Spring Force. Conversely, the object is acted upon by the force that brings back the object to its state of equilibrium of rest.

Magnetic Force:

The motion between electrically charged particles always results in magnetic force. This is the primary force that is active in electric motors and normal magnets.

Electric Force

In order to create an electric force, you'd need to charge two particles and bring them in proximity to one another. Usually, they'll either repel or attract one another. This attraction or repulsion is known as an electric force.

Cheers

You might be interested in

PLZZZZ HELP??

ludmilkaskok [199]

The answer would be A, as B refers to conduction and C and D refer to radiation. Convection is the transfer of different temperature currents, i.e, A

3 0

3 years ago

Read 2 more answers

At which position is the LOWEST potential energy?

Ad libitum [116K]

The answer is position 3, because it is at its lowest point.

Potential Energy is “stored energy.” It is energy that is ready to be converted or released as another type of energy. We most often think of potential energy as gravitational potential energy. When objects are higher up, they are ready to fall back down. When you stretch an object and it has a tendency to return to its original shape, it is said to have elastic potential energy. Chemical potential energy is the stored energy in a substance’s chemical structure that can be released in a chemical reaction or as heat.

7 0

3 years ago

Which action involves the most efficient conversion of electrical energy into thermal energy?

azamat

Answer:

the answer is option d

because in oven which works through electricity ,it will bake the cake (heat energy is produced)

7 0

4 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time t (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time t (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the velocity at t = 8.00 s. Get the speed by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago

A crime suspect fled the scene when police officers entered the building. How far could he run in 10 minutes if he can run 5 mil

emmainna [20.7K]

(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = 5/6 mile

(5/6 mile) x (1,760 yd/mile) = 1,466 and 2/3 yards

3 0

3 years ago