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3 years ago

What is 300N divided by 25 kg?

Physics

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

Explanation:

it is acceleration because we know that

F=ma

a=F/m

a=300N/25 kg

a=12 m/s^2

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IceJOKER [234]

It helps because it's being transported blah blah whatever the last person said when you first asked this question

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3 years ago

The terms intrusive and extrusive are used to describe which one of the three rock groups

tino4ka555 [31]

Answer:

IGNEOUS ROCKS

Explanation: Igneous rocks are those rocks that solidify from magma.

Igneous rock is divided into two ,they are:

1. Intrusive

Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.

2 Extrusive igneous rock is also known as known as volcanic rocks

8 0

3 years ago

How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?

lukranit [14]

<span>work =V*Q =12*50*10^-6

The total work done will be equal to

work = V.Q

which means

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>

8 0

2 years ago

A 295-kg object and a 595-kg object are separated by 4.10 m.

kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

by putting the values

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

$F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}$

F₁₃=2.94 x 10⁻⁷ N

The force between the mass m₂ and m₃

by putting the values

$F_{23}=\dfrac{Gm_2m_3}{r^2}$

$F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}$

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

Lest take at distance x from mass m₂ net force is zero.

$F_{23}=\dfrac{Gm_2m_3}{x^2}$

$F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}$

Form above two equation

$\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}$

$\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}$

$\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}$

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0

2 years ago

awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall

pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

$KE = 6.66 J$

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

$m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}$

now plug in all values in it

$1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22$

so we have

$3.01 = 0.375v_1 + 0.927v_2$

similarly in Y direction we have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}$

now plug in all values in it

$0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22$

so we have

$0 = 0.927v_1 - 0.375v_2$

$v_2 = 2.47 v_1$

now from 1st equation we have

$3.01 = 0.375 v_1 + 0.927(2.47 v_1)$

$v_1 = 1.13 m/s$

$v_2 = 2.78 m/s$

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

$KE = \frac{1}{2}(1.47)(3.01^2)$

$KE = 6.66 J$

5 0

3 years ago