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3 years ago

What is the current in a circuit with a 9.0 V battery and a resistance of 36.0 Ω?

Physics

2 answers:

ddd [48]3 years ago

5 0

Current=voltage/resistance= 9.0 V/36.0Ω=1/4=0.25 A

Firlakuza [10]3 years ago

4 0

I = V / R = 9v / 36ohms = 1/4 Ampere

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A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: Drag Force is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the Terminal Velocity.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

$F_{G}=F_{D}$

$mg=\frac{1}{2}C \rho Av_{T}^{2}$

$v_{T}=\sqrt{\frac{2mg}{\rho CA} }$

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

$v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 9.09

The 95kg skydiver's terminal velocity is

$v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

$t=\frac{4750}{9.09}$

t = 522.5

95 kg at 12.3 m/s:

$t=\frac{4750}{12.3}$

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0

3 years ago

What is the equation for potential energy?

Mrrafil [7]

You can calculate potential energy by:
U = m.g.h

Where, U = potential energy
m = mass
g = acceleration due to gravity
h = height

Hope this helps!

7 0

3 years ago

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A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

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3 years ago

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kiruha [24]

12 1/2 i would assume

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Alexxandr [17]

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