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3 years ago

A famous physics tale is about a rich man who was found dead. He was

1 answer:

7 0

<h2>Answer: C) He could have thrown the bag of money sideways, creating a horizontal reaction force on himself.</h2>

Explanation:

According to Newton's third law of motion, when two bodies interact between them, appear equal forces and opposite senses in each of them.

To understand it better:

Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude and direction but in the opposite direction on the first.

So, if the rich man had pushed the bag of money horizontally opposite of where he was, he could have saved himself.

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Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th

aev [14]

Answer:

A) G = m³/kg.s²

B) E = kg.m²/s²

Explanation:

The given relation is:

F = Gm₁m₂/r²

where, the units of all variables are:

F = Force = kg.m/s²

m₁ = m₂ = mass = kg

r = distance = m

G = Gravitational Constant = ?

Therefore,

kg.m/s² = G(kg)(kg)/m²

(kg.m/s²)(m²/kg²) = G

<u></u>

The given equation is:

E = mc²

where, the units of all variables are:

m = mass = kg

c = speed = m/s

E = Energy = ?

Therefore,

E = (kg)(m/s)²

This is the correct answer, which is not present in any option.

4 0

4 years ago

In order for a wave to occur, there must be a ______________ in the medium.

natta225 [31]

I believe the answer is “amplitude”.

3 0

4 years ago

You are dragging a block on a surface with friction at a steady speed of 2 m/s and exert a force of 5 N to do so. What is the fo

m_a_m_a [10]

Answer:

Friction = 5 N

Explanation:

As we know that block is moving at constant speed

So the acceleration of the block is zero

So we will have

$F_{net} = 0$

for net force to be zero

Force exerted on the object by external system must be counter balanced by the force of friction

So we have

$F_{ex} = F_f$

so we have

$F_f = 5 N$

6 0

4 years ago

A horse travels 80 meters south in 20 seconds. What is its average velocity?

AURORKA [14]

Average velocity is a vector unit (i.e. includes magnitude <em>and </em>direction) calculated by working out distance ÷ time:

80 metres ÷ 20 seconds = 4 metres/seconds (m/s)

Therefore, your final answer is C. 4 m/s south.

7 0

3 years ago

Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E

Leviafan [203]

We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
$\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}$
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2$
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
$mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\$
Because both carts have the same mass we can cancel those out:
$3v=v_A+2v_B$
Now we have our system of equation that we have to solve:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B$
Part A
We need to solve our system for $v_a$ . We will solve second equation for $v_b$ and then plug that in the first equation.
$3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}$
Now we have to plug this in the first equation:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
We will multiply the first equation with 2 and divide by m:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
Now we plug in the second equation into first one:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0$
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
$a=1\\
b=-6v\\
c=v^2-\frac{E}{m}$
Solutions are:
$v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}$
Part B
We do the same thing here, but we must express $v_a$ from momentum equation:
$3v=v_A+2v_B\\
v_A=3v-2v_B$
Now we plug this into our energy conservation equation:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\
3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\
3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\
3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\
3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\
3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0
$
Again we end up with quadratic equation. Coefficients are:
$a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}$
Solutions are:
$v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}$

8 0

3 years ago