Ask question

Ask question

All categories

3 years ago

7

You are heating an iron to iron the shirt you're going to wear to school tomorrow. Which of the following best describes the tra

nsformation
of energy in this example?
Electrical to radiant
Thermal to electrical
Chemical to radiant
Electrical to thermal
2.
3
4 5 6 7
8
9
10
Next

1 answer:

telo118 [61]3 years ago

6 0

Answer:

electrical to the thermal

Explanation:

You might be interested in

A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium. True or

Jet001 [13]

Answer;

The above statement is true.

-A photograph is taken by letting light fall on a light-sensitive medium, which then records the image onto that medium.

Explanation;

-A photograph is created when light is allowed to fall on a light-sensitive medium. The pattern of light creates an image that is recorded by the photographic device. How light or dark a photograph is depends on how much light was allowed to fall on the light-sensitive medium.

-A camera is a light-tight box that contains a light-sensitive material or device and a way of letting in a desired amount of light at particular times to create an image on the light-sensitive material.

6 0

3 years ago

HELP!!!!!! Need Help !!!

denis23 [38]

Answer:

its y

Explanation:

3 0

2 years ago

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Marianna [84]

I found the answer for you if u need any help ask anytime!

3 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

2 years ago

Match each type of wave to the way it moves. transverse wave longitudinal wave electromagnetic wave back and forth at right angl

marta [7]

Transverse wave = at right angles to the direction of the motion of the wave

in transverse wave medium particles will move perpendicular to the direction of motion of medium particles

they all are perpendicular to wave always

Longitudinal wave = back and forth in the direction of the motion of the wave

in longitudinal waves medium particles will move in the direction of wave and the motion is always in back and forth type

electromagnetic wave = alternating waves moving at right angles to each other

electromagnetic waves are combination of electric field and magnetic field which oscillates perpendicular to wave as well as they are perpendicular to each other

8 0

3 years ago

Read 2 more answers