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3 years ago

Which two statements are true about energy transformations

Physics

2 answers:

Lorico [155]3 years ago

8 0

Answer:

First statement is that the total mechanical energy in a close system is always constant.

Second statement is that energy can change from one form to another. e. g from potential energy to kinetic energy

Explanation:

First statement is that the total mechanical energy in closed system is always constant.

Second statement is that energy can change from one form to another.e.g potential energy to kinetic energy.

inn [45]3 years ago

3 0

Hey luv, what’s the two statements?

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The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space

Svet_ta [14]

D not 100% sure though.

4 0

3 years ago

You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double

dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

$\tan\theta=\dfrac{y}{D}$

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

$\tan\theta=\dfrac{1.52}{120}$

$\theta=\tan^{-1}\dfrac{1.52}{120}$

$\theta=0.725$

We need to calculate the wavelength

Using formula of wavelength

$d\sin\theta=n\lambda$

Put the value into the formula

$0.025\times\sin0.725=5\times\lambda$

$\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}$

$\lambda=6.326\times10^{-7}\ m$

$\lambda=633\ nm$

Hence, The wavelength of the light is 633 nm.

4 0

3 years ago

Being too hot or too cold is an example of an internal distraction

otez555 [7]

Is this true or false?

4 0

3 years ago

Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water a

PolarNik [594]

Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

Explanation:

(a)

$L_{f}$ = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

$Q_{f}$ = Energy required to melt the ice

Energy required to melt the ice is given as

$Q_{f}$ = m $L_{f}$

$Q_{f}$ = (1.63) (3.33 x 10⁵)

$Q_{f}$ = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q = Amount of energy remaining to raise the temperature

Using conservation of energy

E = $Q_{f}$ + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

T = 45 °C

5 0

3 years ago

In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls

sweet-ann [11.9K]

Answer:

Acceleration= $0.5m/s^2$

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration= $a=\frac{F}{m}$

Using the formula

Acceleration of box= $\frac{20}{40}=0.5m/s^2$

The acceleration of the box= $0.5m/s^2$

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

$a=\frac{30}{40}=\frac{3}{4} ms^{-2}$

$v^2-u^2=2as$

Substitute the values

$v^2-0=2\times \frac{3}{4}\times 0.3=0.45$

$v^2=0.45$

$v=\sqrt{0.45}=0.67m/s$

Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s

4 0

3 years ago