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2 years ago

Why is Energy not created nor destroyed? (Need the answer ASAP!!)

1 answer:

eduard2 years ago

4 0

Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the surroundings outside the system. In other words, energy cannot be created or destroyed.

You might be interested in

Which statement is true?

miv72 [106K]

B might be the correct answer

8 0

2 years ago

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

Yuri [45]

The total power emitted by an object via radiation is:
$P=A\epsilon \sigma T^4$
where:
A is the surface of the object (in our problem, $A=1.25 m^2$
$\epsilon$ is the emissivity of the object (in our problem, $\epsilon=1$ )
$\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is $T=100^{\circ} C=373 K$

Substituting these values, we find the power emitted by radiation:
$P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW$
So, the correct answer is D.

6 0

3 years ago

a 5.00 kg plane is speeding up on the ground with an applied force of 706 N. If the net force is 450 N, what would be the force

loris [4]

Answer:

The force due to air resistance is 256 N.

Explanation:

Given;

mass of the plane, m = 5 kg

applied force on the plane, Fa = 706 N

the net force on the plane, ∑F= 450 N

Let the force due to air resistance = Fr

The net force on the plane is given as;

Net force = applied force - force due to air resistance

∑F = Fa - Fr

Fr = Fa - ∑F

Fr = 706 - 450

Fr = 256 N.

Therefore, the force due to air resistance is 256 N.

5 0

3 years ago

A 6.50 × 10–3 m2 piston compresses gas in a cylinder with a surface area of 9.75 × 10–2 m2. What is the force on the cylinder wa

vodomira [7]

Yes it is 2cm and 1 quarter

8 0

2 years ago

A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte

Afina-wow [57]

Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

t = 3.2 / √( 1 - 0.36 )

t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

6 0

3 years ago