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irga5000 [103]
3 years ago
13

assuming that there are no outside forces, what will happen if two neutral atoms are placed 1 M apart in a vacuum. a)the atoms w

ill orbit each other b)the atoms will remain stationary c) the atoms will move tword each other d) the atoms will move away from each other.
Physics
1 answer:
timurjin [86]3 years ago
7 0
D the atoms will move tword each other hope this helps
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Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms​
aev [14]

Answer:

The mass of 0.02 m³ of gold is 386 kilograms

Explanation:

Given:

The density of the gold = 19300 kg/m³.

The volume of gold = 0.02 m³

To Find:

The mass of gold = ?

Solution:

We know that density is mass divided per unit volume.

Thus mathematically

Density = \frac{mass}{volume}Density=

volume

mass

Rewriting in terms of mass ,

Mass = density * volume

On substituting the known values

Mass = 19300 kg/m³ * 0.02 m³

Mass = 386 kilograms

Learn more about Mass and Density:

Mass=?,volume=190,density=4

Mass 350 kg volume 175 density ans

This is not my answer I copied it but hope it helps:)

5 0
3 years ago
an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.​
Brut [27]

Answer:

0.5m/s2

Explanation:

acceleration= change in velocity/time taken

= v - u/ t

= 10-5/10

=5/10

= 0.5m/s2

6 0
3 years ago
Use the concepts of kinetic energy and potential energy to describe the motion of a child on a swing. Why does the child need a
andrew-mc [135]
When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.

The child may need a push from time to time because friction with the air causes loss of energy.
4 0
2 years ago
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By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 301 m/s to 343
garri49 [273]

Answer:

29.8 %

Explanation:

7 0
3 years ago
A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c
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First establish the summation of the forces acting int the ladder

Forces in the x direction Fx = 0 = force of friction (Ff) – normal force in the wall(n2)

Forces in the y direction Fy =0 = normal force in floor (n1) – (12*9.81) –( 60*9.81)

So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

5/ 3 = h/ 0.844

H = 1.4 m can the 60 kg person climb berfore the ladder will slip

7 0
3 years ago
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