Which pair of quantities represent scalar quantities, and how are they different from vector quantities?

The force supplied by a spring depends on what?

Morgarella [4.7K]

The force and motion

8 0

3 years ago

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

a

The mass is $m_2 =21.75*10^{-27} \ kg$

b

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

So

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

So

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago

A car has a mass of 900 kilograms and has a force of 4500 to the left what is the acceleration

pychu [463]

Answer: A- A truck has a mass of 900 kilograms.

If the velocity of the truck is 20 m/s, what is its momentum?

1- 180000 kg·m/s

2-0 kg·m/s

3-9000 kg·m/s

4-18000 kg·m/s

3 0

2 years ago

Justin is riding his bike up Dunmore Hill. The effort force from his feet on his pedals is 1160N.

Schach [20]

Explanation:

the answer is 1835N that how best I can help

3 0

2 years ago

Two cyclists start at the same point, but travel along two straight streets in different directions. if they continue, how many

Mumz [18]

Since they are travelling on two different straight directions, therefore their paths will NEVER cross again. This because straight lines will keep on going forever on the same direction therefore there is 0 possibility for two straight lines to paths cross again.

Answer:

<span>zero or never</span>

8 0

3 years ago