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3 years ago

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What’s the difference between gravitational force & gravitational field strength?

2 answers:

Lelu [443]3 years ago

7 0

what’s the difference between gravitational force & gravitational field strength?

Zepler [3.9K]3 years ago

5 0

Answer:

Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.

Explanation:

hope it helps u

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Two 8V batteries are set up with four light bulbs in a complex circuit. Light bulb A has a resistance of 4.00 Ω, light bulb B ha

Neko [114]

Answer:

Total resistance : 12.4

Total Current : 0.645

Explanation:

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3 years ago

A 238 N force is applied to a 25 kg object. What is the object's acceleration?

Ganezh [65]

Answer:

<h2>9.52 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{238}{25} \\ = 9.52$

We have the final answer as

<h3>9.52 m/s²</h3>

Hope this helps you

7 0

3 years ago

According to Newton's second law of motion, if we we have a rigid, unchanging mass and we observe it accelerating, what must be

faltersainse [42]

A. a force is being applied to the mass

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3 years ago

Read 2 more answers

How long would it take you too run the 100m sprint if from rest you accelerate at 2m/s/s for the whole race?

fredd [130]

Answer:

10 s

Explanation:

Given:

Δx = 100 m

v₀ = 0 m/s

a = 2 m/s²

Find: t

Δx = v₀ t + ½ at²

100 m = (0 m/s) t + ½ (2 m/s²) t²

t = 10 s

4 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

3 years ago