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Tomtit [17]
3 years ago
11

What is the meaning of mass​

Physics
2 answers:
ruslelena [56]3 years ago
7 0

Answer:

E = M c^2     Einstein's definition of mass

M = E / c^2 = E / (3 * 10^8^2) =  (kg m^2 / s^2) /  (m / s)^2 = kg

E and M are equivalent definitions

ValentinkaMS [17]3 years ago
3 0

Answer:

anything that takes up space.

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An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in
SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

h=-16t^2+v_ot

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

-16t^2+250t=500

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0
4 years ago
Read 2 more answers
Helppppppppppppppppppppppppp
andrew11 [14]

Answer:

value of x=35°

y=72.5°

hope it helps you

make me brainliest plz

6 0
3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
A potential energy function is given by u(x)=(3.00n)xâ(1.00n/m2)x3. at what position or positions is the force equal to zero?
qwelly [4]

I believe the correct form of the energy function is:

u (x) = (3.00 N) x + (1.00 N / m^2) x^3

or in simpler terms without the units:

u (x) = 3 x + x^3

Since the highest degree is power of 3, therefore there are two roots or solutions of the equation.

 

Since we are to find for the positions x in which the force equal to zero, u (x) = 0, therefore:

3 x + x^3 = u (x)

3 x + x^3 = 0

Taking out x:

x (3 + x^2) = 0

So one of the factors is x = 0.

 

Finding for the other two factors, we divide the two sides by x and giving us:

x^2 + 3 = 0

x^2 = - 3

x = sqrt (- 3)

x = - 1.732 i, 1.732 i

 

The other two roots are imaginary therefore the force is only equal to zero when the position is also zero.

 

Answer:

x = 0

5 0
4 years ago
Which of the following is an example of a physical change but not a chemical change?
vazorg [7]

A water pipe freezes and cracks on a cold night. <em> (D)</em>

There are three physical changes going on in this scenario:

1).  The air gets cold and dark at night.

2).  The water in the pipe freezes and expands.

3).  The pipe cracks.

There are no chemical changes in the description.

7 0
3 years ago
Read 2 more answers
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