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bulgar [2K]
3 years ago
11

10 points!! If you help!!!

Physics
1 answer:
marta [7]3 years ago
4 0

For Mass

K.E = (1/2*mv^2)

Explanation:

Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

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25.5 m/s in 5.75 s. What is the acceleration?
anzhelika [568]
Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)

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7 0
3 years ago
Compared to Earth's moon, the moons of Mars are
Mice21 [21]
I think is b I think
7 0
3 years ago
Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri
Bess [88]

Answer: 0.75\ cm

Explanation:

Given

Wavelength of light \lambda=500\ nm

Screen is D=1\ m away

Distance between two adjacent bright fringe is \Delta y=\dfrac{\lambda D}{d}

When same experiment done in water, wavelength reduce to \dfrac{\lambda }{\mu}

So, the distance between the two adjacent bright fringe is \Delta y'=\dfrac{\lambda D}{\mu d}

Keeping other factor same, distance becomes

\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm

3 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

(c) 45 °C

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
Please look at the photo and give me the answer :)))
san4es73 [151]

Answer:

c

Explanation:

efficiency is measured in % hence 0.02% is smaller than 20%

5 0
2 years ago
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