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2 years ago

Given that on Earth, gravity causes an acceleration of 9.8 m/s2, what is an acceleration of 7 g?

Physics

1 answer:

zzz [600]2 years ago

7 0

Answer:

68.6 m/s^2

Explanation:

1 g = 9.8 m/s^2

7 g × 9.8m/s^2 = 68.6

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Choice-C is a correct statement.

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Which is the best definition of a parallel

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A parallel circuit exists when an electric charge flows in more than one path best describes it.

<h3>What is a Parallel circuit?</h3>

This type of circuit has branches in which the current divides and only part of it flows through any of the branch.

Parallel circuit having more than one branch therefore means that electric charge will flow in more than one path thereby making option A the most appropriate choice.

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2 years ago

An object with a mass 4.0 kg has a momentum of 64 kgm/s . How fast is the object traveling ?

Anastasy [175]

Answer:

The object will travel at the speed of 16 m/s.

Explanation:

Given

Mass m = 4.0 kg

Momentum p = 64 kgm/s

To determine

How fast is the object traveling?

<u>Important Tip:</u>

The product of the mass and velocity of an object — momentum.

Using the formula

$p = mv$

where

m = mass

v = velcity

p = momentum

Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula

$p = mv$

$64\:=\:4\times v$

switch the equation

$\:4\times \:v\:=64$

divide both sides by 4

$\frac{4v}{4}=\frac{64}{4}$

simplify

$v=16$ m/s

Therefore, the object will travel at the speed of 16 m/s.

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2 years ago

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

podryga [215]

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

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