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Vladimir [108]
3 years ago
10

The specific heat of water is 4190 j/(kg*k). suppose you put 1 kg of water (a bit over 4 cups) into a microwave that can deliver

1000 w of power to the water when it is running at full power. roughly how long will you need to run the microwave at full power in order to raise the temperature of the water by 10°c?
Physics
1 answer:
VladimirAG [237]3 years ago
4 0
Given:
P = 1000 W, power input
c = 4190 J/(kg-K), the specific heat of water
m = 1 kg, mass of water
ΔT = 10 °C = 10 K, temperature rise.

Let t = time required to raise the temperature f the water.
Then
P*t = m*c*ΔT
(1000 J/s)*(t s) = (1 kg)*(4190 J/(kg-K)*(10 K)
1000t = 41900
t = 41.9 s

Answer: 41.9 s or 42 s (approximately)

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A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
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The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

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  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0
2 years ago
In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?
agasfer [191]

Answer:

The frequency of the photon is 3.069\times10^{14}\ Hz.

Explanation:

Given that,

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Using relation of energy

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Where, \Delta E =  energy spacing

4h\nu-2h\nu=4.07\times10^{-19}

\nu=\dfrac{4.07\times10^{-19}}{2h}

Put the value of h into the formula

\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}

\nu=3.069\times10^{14}\ Hz

Hence, The frequency of the photon is 3.069\times10^{14}\ Hz.

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3 years ago
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Answer:

7. Your answer is correct dear, just add the unit

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Explanation:

I hope that help's

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