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sesenic [268]
3 years ago
13

if i release one steel ball from the top of a ramp and the other ball from the 40cm mark will they have the same velocity?

Physics
1 answer:
k0ka [10]3 years ago
5 0

Answer:

Yes.

Explanation:

Assuming that the balls weigh the same, then they would have the same velocity. Velocity is the term used to describe the speed at which an object is traveling.

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Please help and show workings. <br>​
Mnenie [13.5K]

Effective resistance = 2+1.5 = 3.5 ohms

Current in the circuit = 1.2 A

Lost volts in the battery = 1.8V

Current in one of the 3 ohms resistors = 0.6 A

I hope it helps

3 0
3 years ago
Of all fresh water available how much of it is ready for use by humans
Vlad1618 [11]
For a human. the human supply water usage is 3%. ONLY 3 PERCENT IS DRINKABLE TO HUMANS
4 0
3 years ago
4. An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5
Mazyrski [523]

Answer:

11.87 ms⁻¹

Explanation:

You can use the kinematic equation

v² = u² + 2as

Where v = final velocity

          u = initial velocity

          a = acceleration

          s = displacement

v² = 9² + 2×1.5×20

So you get v = 11.87 ms⁻¹

7 0
3 years ago
humming bird flits from flower to flower. Over a period of 10 minutes, it travels 40 meters. The average speed of the humming bi
Veronika [31]

As we know that average speed is given by the ratio of total distance covered and total time of motion so we will have

v_{avg} = \frac{distance }{time}

now we have

distance covered = 40 meter

time taken by the bird = 10 minutes

now in order to find the average speed of the bird

v_{avg} = \frac{40}{10}

v_{avg} = 4 meter/min

so the average speed of the bird is 4 m/min

4 0
4 years ago
In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ
grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

I = \frac{1}{2}ε₀cE_{max²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E_{max is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E_{max  )

E_{max = 350/f kV/m

given that frequency of 60 Hz, we substitute

E_{max = 350/60 kV/m

E_{max = 5.83333 kV/m

E_{max = 5.83333 kV/m × ( \frac{1000 V/m}{1 kV/m} )

E_{max = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

I = \frac{1}{2}ε₀cE_{max²

I = \frac{1}{2} × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

I = 45 × 10³ W/m²

I = 45 × 10³ W/m² × ( \frac{1 kW/m^2}{10^3W/m^2} )

I = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0
3 years ago
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