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xxMikexx [17]
3 years ago
14

The mass of 2 cm3 of gold is 38.6 grams. find the density of the gold. Could you please help me??

Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Density = mass/ volume (so here this is how you would solve the problem)

<span>D = 38.6 g/ 2 cm3 (first step)</span>

<span>D= 19.3 g/cm3  ( Do math and then you would get this)</span>

<span>
</span>

<span>Hope this helps!! :) </span>


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A motor requires 400 joules of energy to lift a 5.0 kg mass 2.0 meters. Calculate the efficiency of this motor.
Sauron [17]

Answer:

25%

Explanation:

use F=mg

then use the answer you get from that and plug it into W=Fxh

take that answer and divide it by 400 J and multiply by 100

round to sig figs

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3 years ago
As potential and kinetic energy increase what happens to mechanical energy?
Tom [10]

Answer:

a

Explanation:

7 0
3 years ago
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A convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of
OLga [1]

Answer:

<h2>36cm from the surface</h2>

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

\dfrac{n_2}{v} = \dfrac{n_1}{u}=  \dfrac{n_2-n_1}{R}

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = \infty (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

<em>focus point of the light that is incident and parallel to the central axis (v)</em>

<em />

Substituting this values into the given formula we will have;

\dfrac{1.5}{v} - \dfrac{1}{\infty}=  \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0=  \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}=  \dfrac{0.5}{12}\\\\

Cross multiply

1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm

Hence  Light incident parallel to the central axis is focused at a point 36cm from the surface

6 0
3 years ago
Find the force between charges of +10.0 C and -50.0 C located 20.0 cm apart
polet [3.4K]

The magnitude of the electrostatic force between the two charges is 1.12\cdot 10^{14} N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

q_1 = +10.0 C

q_2 = -50.0 C

r = 20.0 cm = 0.20 m

Therefore, the force between the charges is

F=(8.99\cdot 10^9) \frac{(10)(-50)}{(0.20)^2}=-1.12\cdot 10^{14} N

where the negative sign means that the force is attractive, since the two charges have opposite sign.

Learn more about electrostatic force here:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
Is an electron cloud bigger than an atom?
Rashid [163]

Answer:

no

Explanation:

An electron is one of the components of an atom, so it cannot be larger than that.

5 0
2 years ago
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