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Otrada [13]
3 years ago
14

How much force is required to accelerate a 9.0-g object at 10000 g's?

Physics
1 answer:
Yuki888 [10]3 years ago
6 0
Hey give us m = 9.0 g = 9.0 x 10-3 kg, and a = 10,000 "g's" = 98000 m/s/s so:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N
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The current through two identical light bulbs connected in series is 0.25 A. The voltage across both bulbs is 110 V. The resista
Rus_ich [418]

The resistance of a single light bulb is 220 ohms per bulb.

<h3>What is Ohm's Law?</h3>

Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.

Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.

E = I x R

The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.

110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.

R = 110/(2*0.25) = 220 ohms

to learn more about Ohms law go to - brainly.com/question/14296509

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2 years ago
State the 3 laws of motion
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Answer:

Explanation:

Newton's first law of motion:

An object in motion stays in motion, and an object at rest stays at rest, until acted upon by an unbalanced force.

Newton's second law:

The net force on an object is equal to its mass times its acceleration.

Newton's third law:

For every action, there is an opposite and equal reaction.

6 0
3 years ago
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DochEvi [55]

Answer:

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6 0
2 years ago
What law states force is dependent on the mass and acceleration of an object
UNO [17]

Answer:

Newton's second law of motion

Explanation:

Newton's second law of motion can be stated  

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5 0
3 years ago
A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag
VARVARA [1.3K]

Answer:

Induced emf, \epsilon=9.79\times 10^7\ volts

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, \omega=3\ rev/s=18.84\ rad/s

The vertical component of the Earth’s magnetic field is, B=6.5\times 10^5\ T

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

\epsilon=\dfrac{1}{2}B\omega l^2

\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2

\epsilon=9.79\times 10^7\ volts

So, the induced emf between the tip of a blade and the hub is \epsilon=9.79\times 10^7\ volts. Hence, this is the required solution.

5 0
3 years ago
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