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Molodets [167]
2 years ago
5

A 69 kg zebra is traveling 7 m/s east. What is the zebra’s momentum? kg-m/s

Physics
1 answer:
Hitman42 [59]2 years ago
6 0
Linear momentum is the product of mass and velocity. In this case, it is simply:
69\times 7 = 483 kg\cdot m/s
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2. What is the resistance expected on a heater that is feed by 480V AC and the Electric Power is 4.6kW? ​
Korvikt [17]

The resistance expected of the heater is 50.1 ohms.

<h3>What is resistance?</h3>

Resistance can be defined as the opposition to the flow of electric current in an electric circuit. The S.I unit of resistance is Ohms (Ω).

To calculate the resistance of the heater, we use the formula below.

<h3>Formula:</h3>
  • R = V²/P............. Equation 1

Where:

  • R = Resistance of the heater
  • P = Power of the heater
  • V =  Voltage supplied to the heater

From the question,

Given:

  • V = 480 V
  • P = 4.6 kW = 4600 W

Substitute these values into equation 1

  • R = (480²)/4600
  • R = 50.1 ohms.

Hence, the resistance expected of the heater is 50.1 ohms.

Learn more about resistance here: brainly.com/question/17563681

3 0
2 years ago
As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th
poizon [28]

Answer:

4.048\times 10^{-7}\ m

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

E = Energy = 4.91\times 10^{-19}\ J

Wavelength ejected is given by

\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m

The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is 4.048\times 10^{-7}\ m

4 0
3 years ago
Would someone plz help a shister outttt
tiny-mole [99]

Yaaaas, do you watch James Charles!?

8 0
3 years ago
A form of oxygen with three atoms of
lesya692 [45]

Answer:

ozone, letter b

ozone is basically an allotrope of oxygen having three atoms instead of two.

5 0
2 years ago
At what rate would heat be lost through the window if you covered it with a 0.750 mm-thick layer of paper (thermal conductivity
igor_vitrenko [27]

The given question is incomplete. The complete question is as follows.

A picture window has dimensions of 1.40 m, 2.50 m and is made of glass 6.00 mm thick. On a winter day, the outside temperature is -15^{o}C, while the inside temperature is a comfortable 21.0^{o}C.

Part A

At what rate is heat being lost through the window by conduction?

Express your answer using three significant figures.

Part B

At what rate would heat be lost through the window if you covered it with a 0.750 mm-thick layer of paper (thermal conductivity 0.0500 W/m?K)?

Express your answer using three significant figures.

Explanation:

Formula for rate of heat transferred through single thick plane of glass is as follows.

       (\frac{Q}{\Delta t)}_{single} = \frac{A(T_{h} - T_{c})}{\frac{L_{ghss}}{K_{ghss}}}

                               = \frac{1.4 \times 2.5 m^{2} \times 36^{o}C}{\frac{6.0 mm}{0.8 W/m K}}

                           = 16.8 \times 10^{3} W

When window is covered by paper then the rate of heat transfer is as follows.

(\frac{Q}{\Delta t)}_{single} = \frac{A(T_{h} - T_{c})}{\frac{L_{ghss}}{K_{ghss}} + \frac{L_{paper}}{K_{paper}}}

                 = \frac{(1.4 \times 2.5 m^{2} \times 36^{o}C}{\frac{6.0 mm}{0.8 W/m K} + \frac{0.75 mm}{0.05 W/m K}}

                 = 5.6 \times 10^{3} W

Thus, we can conclude that heat lost is 5.6 \times 10^{3} W.

3 0
3 years ago
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