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Zinaida [17]
3 years ago
5

Define relative density

Physics
2 answers:
julia-pushkina [17]3 years ago
8 0

Relative density is the ratio of the density of a substance to the density of a standard substance like water or air.

egoroff_w [7]3 years ago
3 0

Hello There!

Relative density is another word for specific gravity. Relative density is comparing the density of one thing to the density of another. Most of the time, you look at the density of water so we look at the density of an object relative to the density of water.

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A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is
HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

5 0
3 years ago
1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
Kazeer [188]
I believe it is meteoroids
6 0
3 years ago
Read 2 more answers
How Many Significant Digits are in 0.0217​
Bumek [7]

Answer:

<em>The number 0.0217 has 3 significant digits</em>

Explanation:

<u>Significant Digits </u>

These are digits that contribute to the significance of the number. Some rules apply to discard the non-significant digits like:

  • Leading zeros
  • Trailing zeros (with exceptions)

Our number is 0.0217 has two leading zeros before the 2 because they only occupy space to indicate the order of magnitude of the number. Only the 2,1,7 are significant digits, thus

The number 0.0217 has 3 significant digits

7 0
3 years ago
If a person (weighs 70kg) jumped of a moving car (at 100km/h) and fell on asphalt what is the amount of force applyied to his bo
jok3333 [9.3K]

Answer:

The amount of force applied to his body is 1944.44 N

<em>The chances of the person dying is very high owing to the high impact force with which the person would experience when he or she lands on the asphalt road due to the jump out of the moving car.</em>

Explanation:

We all know that,

F = Ma where,

F = Force

M = weight of the person

a = acceleration or velocity of the moving car

Therefore;

F = { 70 x (100 x 1000) } / [3600]

= [7 000 000] / 3600

= <u>1944.44 N</u>

6 0
3 years ago
If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ra
igomit [66]

Answer:

\frac{E}{E_c} =3.125

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

E_c=\frac{1}{2} mv^2

The equivalence between mass and energy established by the theory of relativity is given by:

E=mc^2

This formula states that the equivalent energy E can be calculated as the mass m multiplied by the speed of light c squared.

Where c is approximately 3\times 10^{8} m/s

Hence:

E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J

E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J

Therefore,  the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125

4 0
3 years ago
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