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3 years ago

3. As an object’s temperature increases, the ____________________ at which it radiates energy increases.

Physics

2 answers:

Vladimir [108]3 years ago

7 0

Answer:

As an object’s temperature increases, the Rate at which it radiates energy increases.

vredina [299]3 years ago

7 0

An increase in temperature of the object’s, increases the <u>RATE</u> that radiates energy.

<u>Explanation: </u>

The difference in the temperature of the object is the main reason for transfer of heat energy from one to the other. The overall average of kinetic energy produced by the particles that exits in a substance is known to be temperature.

The temperature may be either cold or hot. When there is a rise or increase in an object's temperature, there will be an increase in rate at which it radiates or emits energy. The molecule in the object moves when it gets heated.

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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0

3 years ago

A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp

Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0

2 years ago

two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma

nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force = 3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos $\theta$

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)

35 = √49 x²

x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0

3 years ago

Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so t

alexandr1967 [171]

Answer:

New force, $F'=48\times 10^3\ N$

Explanation:

It is given that,

Force acting between two charged particles, $F=7.5\times 10^2\ N$

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

$F=k\dfrac{q_1q_2}{r^2}$ ............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

$F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}$ ..........(2)

Dividing equation (1) and (2) :

$\dfrac{F}{F'}=\dfrac{1}{64}$

$F'=7.5\times 10^2\ N\times 64$

F' = 48000 N

$F'=48\times 10^3\ N$

Hence, this is the required solution.

8 0

3 years ago

Light is a form of what

mart [117]

Light is a form of energy.
(:

8 0

3 years ago