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3 years ago

Connecting

Physics

1 answer:

svetoff [14.1K]3 years ago

4 0

5. what is the law called that states objects in motion must stay in motion/objects at rest will stay at rest?

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You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2

oee [108]

Answer:

Explanation:

Given

Resistor A has length $L_1$

and Resistor B has Length $L_2$

and Resistance is given by

$R=\frac{\rho L}{A}$

Considering $\rho$ and A to be constant thus

$R_2>R_1$ because $L_2>L_1$

(a)When they are connected in series

As the current in series is same and power is $i^2R$

therefore $P_2>P_1$ as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

$P=\frac{V^2}{R}$

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

8 0

3 years ago

What is the formula to find initial velocity?

Inga [223]

Answer:

s = ut + 1/2at^2

Explanation:

s= distance

u= initial velocity

a= acceleration

7 0

3 years ago

A 25.0kg girl pushes a 50.0kg boy with a force of 100.0N. What is the acceleration of the girl?

MA_775_DIABLO [31]

Answer:

im pretty sure it should be 50.0

6 0

3 years ago

Bài 5: Một electron di chuyển được đoạn đường 1cm, dọc theo một đường sức điện, dưới tác dụng của lực điện trong một điện trường

eimsori [14]

Explanation:

guy thy guy thy guy thy guy thy guy thy

7 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago