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Softa [21]
3 years ago
9

What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s

Physics
1 answer:
Doss [256]3 years ago
7 0

The kinetic energy of any moving object is

                           (1/2) (mass) (speed²) .

For the object you described, that's

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
______________________________

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Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential
kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately \rm 2.1 \; V.

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

<h3>Part A</h3>

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of R_1 + R_2 = 50 + 90 = 140\; \Omega. The voltage across the two resistor, combined, is equal to \rm 6\; V. Hence by Ohm's Law, the current through the circuit will be equal to \rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A.

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to \rm \dfrac{3}{70}\; A. Apply Ohm's Law (again) to find the voltage across R1:

V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V.

<h3>Part B</h3>

Since the equivalent resistance is equal to R_1 + R_2, when the value of R_1 decreases, the equivalent resistance will also decrease. By Ohm's Law, I = \dfrac{V}{R}. When the value of the denominator ( decreases, the value of the quotient, I the current through the circuit, will increase.

<h3>Part C</h3>

Keep in mind that if two resistors are connected in series,

I(R_1) = I(\text{Circuit}) = I(R_2).

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0
3 years ago
A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o
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Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = \omega=7.27\times 10^{-5} rad/s

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s

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6 0
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rusak2 [61]

Answer:

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