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3 years ago

Identify the energy transformations that take place in an electrical fan heater

Physics

2 answers:

natulia [17]3 years ago

4 0

Answer:

Here the electrical energy is transformed into thermal energy

Explanation:

When electrical heater with fan is used to transmit the energy then in this case the energy supplied by the heater to the air molecules which is then forced to move by convection mode of heat transfer.

Here the transformation of energy is given in the form of electrical energy to thermal energy.

this thermal energy is absorbed by the air molecules which are then forced or pushed away by the fan

This type of energy transfer is known as thermal convection of heat.

So here the energy transformation is as following

Electrical energy to Thermal energy then by fan it is kinetic energy of molecules

Sergio [31]3 years ago

3 0

The energy goes from electric energy and gets converted into thermal energy.

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

3 years ago

Two samples of water are mixed together.

horrorfan [7]

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x) Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x) Remove the brackets.

0.5x = 150 - 1.5x Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x Combine like terms

2x = 150 Divide by 2

x = 75

Answer

6 0

3 years ago

Beta emission, alpha emission, positron emission, and electron capture are processes that result in a change in atomic number. W

Tema [17]

Alpha emission is the process results in a change in mass number. Option B is correct.

<h3>What is mass number?</h3>

The total number of protons and neutrons in an atomic nucleus is known as the mass number, often known as the atomic mass number or nucleon number.

It's about the same as the atom's atomic mass, expressed in atomic mass units.

The alpha particle is a helium nucleus with two protons and two neutrons in an alpha decay or alpha emission. The number of protons and neutrons is reduced by two as a result of this action.

The quantity of protons and neutrons is affected by gamma emission descent. Also, while electron capture has no effect on the number of neutrons, it does raise the 1 also number of protons by one.

Alpha emission is the process results in a change in mass number.

Hence option B is correct.

To learn more about the mass number, refer:

brainly.com/question/4408975

#SPJ1

5 0

2 years ago

How does light and heat energy from the sun reach earth?

defon

Explanation:

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2 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

3 years ago