Please answer quick!!

1. A particle has a mass of 10 kg and a velocity of 5 m/s. What is the momentum of the particle?

Aleksandr [31]

✰ Concept :-

⠀

We know that momentum is the product of mass and velocity. Thus, by the following definition, we get :-

⠀

$\qquad \maltese \: \underline{ \boxed{ \purple{ \sf Momentum = Mass \times Velocity }}} \: \star$

⠀

✰ Given Information :-

⠀

A particle has mass 10 kg
Moving with velocity 5 m/s

⠀

✰ To Find :-

⠀

The momentum of the particle

⠀

✰ Solution :-

⠀

Using the formula from the concept section, we get :-

⠀

$\sf \longrightarrow Momentum =10 \: kg \times 5 \: {m}^{ - 1} ~~~ \\ \\ \\\sf \longrightarrow Momentum = {50~kg \: ms}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow \underline{ \boxed{ \frak{ \orange{Momentum = {50 ~kg\: ms}^{ - 1} }}}} \: \star \: \: \: \: \: \: \\ \\$

Thus, the momentum of the particle is 50 kg m/s.

⠀

$\underline{ \rule{230pt}{2pt}} \\ \\$

8 0

3 years ago

A big box of sausages (30 kg) are lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spe

morpeh [17]

Answer:

The work done by gravity moving the box up is -514.5 Joules.

Explanation:

The gravity acts in the direction of the displacement, which in this case is negative (ground - top shelf), and so the resulting work is negative. The work is calculated as:

$W_g = mg\Delta h = 30kg\cdot 9.8\frac{m}{s^2}\cdot(0-1.75)m=-514.5J$

The work done by gravity moving the box up is -514.5 Joules.

4 0

3 years ago

A truck covers 40.0 m in 7.10 s while uniformly slowing down to a final velocity of 2.05 m/s.

Alona [7]

Answer:

(A) Original speed= 9.22 m/s

(B) Acceleration= -1.0099 m\s^2

Explanation:

A truck covers 40m in 7.10 secs

The truck slows down at a uniform velocity of 2.05 m/s

(A) The original speed can be calculated as follows

Vo= 2(40)/7.10 - 2.05

= 80/7.10 - 2.05

= 11.2676 - 2.05

= 9.22m/s

(B) The acceleration can be calculated as follows

a= Vf-Vo/t

= 2.05-9.22/7.10

= -7.17/7.10

= -1.0099m/s^2

5 0

4 years ago

An electric dipole of dipole moment 'p' is placed in the position of stable equilibrium in a uniform field of intensity 'E'. The

larisa86 [58]

Answer:

Torque by electric dipole = pEcos thita

5 0

4 years ago

A 40-w lightbulb connected to a 120-v source experiences a voltage surge that produces 132 v for a moment. by what percentage do

andrey2020 [161]

R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%

3 0

3 years ago