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lianna [129]
3 years ago
8

Cubes are three-dimensional square shapes that have equal sides. What is the density of a cube that has a mass of 12.6 g and a m

easured side length of 4.1 cm? (Density: D = m/v )
Physics
1 answer:
sweet-ann [11.9K]3 years ago
6 0
 I got 3.07cm³. Its just 12.6 divided by 4.1
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Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?
MakcuM [25]
The angle of refraction would be further less 
3 0
3 years ago
A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
muminat

Answer:

The speed of the plank relative to the ice is:

v_{p}=-0.33\: m/s

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

m_{g}v_{g}+m_{p}v_{p}=0 (1)

Where:

  • m(g) is the mass of the girl
  • m(p) is the mass of the plank
  • v(g) is the speed of the girl
  • v(p) is the speed of the plank

Now, as we have relative velocities, we have:

v_{g/b}=v_{g}-v_{p}=1.55 \: m/s (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

45v_{g}+168v_{p}=0

v_{g}-v_{p}=1.55

v_{p}=-0.33\: m/s

I hope it helps you!      

8 0
3 years ago
S waves are also known as
qaws [65]
B. secondary waves aka shear waves          
5 0
3 years ago
Read 2 more answers
a 60 kg woman in a elevator is accelerating upward at a rate of 1.2 m/s2. What is the gravitational force acting upon the woman?
Flura [38]

The gravitational force acting upon the woman is equal to <u>-588.6N</u>

Why?

To solve the problem, we need to consider that two forces are acting upon the woman, the first one is related to her weight and the second one is related to the acceleration of the elevator.

Gravitational force acting upon the woman:

Force=m*gravityacceleration\\\\Force=60kg*-9.81\frac{m}{s^{2}}=-588.6N

Hence, we have that the gravitational force acting upon the woman is equal to -588.6N.

Have a nice day!

5 0
3 years ago
Pre-laboratory Assignment: Experiment 20 Reflection and Refraction of Light 1. When light is incident on a reflective surface, w
shusha [124]

Answer:

1) ngle of incidence and reflection are equal,  light carries does not change

2) the angle of this line with respect to the surface is 90º

3) protractor

4)    n₂  sin θ₂ = n_1 sin θ₁,  light ray must have a greater angle than the incident ray ,

Explanation:

1) When light falls on a reflective surface, the angle of incidence and reflection are equal and as it travels in the same medium, the speed that the light carries does not change

2) The normal is a line perpendicular to the point of incidence of light, so the angle of this line with respect to the surface is 90º

3) Angles are measured with a protractor

4) When light passes from one medium to another, the speed of the ray changes due to the difference in the refractive index in each medium, due to this change in speed the transmitted light ray must have a greater angle than the incident ray , since the speed increases as the density of the medium decreases

           \frac{sin \theta _2}{ sin \theta_1} = \frac{v_2}{v_1}

          \frac{c}{v_2} \  sin \theta_2 = \frac{c}{v_1}  \ sin \theta_1

          n₂  sin θ₂ = n_1 sin θ₁

6 0
3 years ago
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