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3 years ago

What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.

Physics

1 answer:

Stolb23 [73]3 years ago

5 0

Answer:

$Momentum(p) = 20kgm/s$

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

$Momentum(p) = Mass(m) * Velocity(v)$

$p = mv$

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

$Momentum(p) = 10kg * 2m/s$

$Momentum(p) = 10 * 2 * kg * m/s$

$Momentum(p) = 20kgm/s$

Hence, the momentum of the man is $20kgm/s$

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A cat has a mass of 4.6 kg. What is its weight on Earth's surface?

andreyandreev [35.5K]

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Earth's gravity is approximately 9.81

weight = mass x gravity

weight = 4.6 x 9.81

weight = 45.126

Answer is B. 45N

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

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4 0

3 years ago

Read 2 more answers

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

2 years ago

The angle of incidence of a light beam in air onto a reflecting surface is continuously variable. The reflected ray is found to

Mariana [72]

Answer:

n = 2.0686

Explanation:

When an unpolarized ray of light is reflected on a surface, the reflected ray is partially polarized, complete polarization occurs when it is true that between the transmitted and reflected ray one has 90, the relationship is

n = so tea

let's calculate

n = tan 64.2

n = 2.0686

8 0

3 years ago

Helpppp!!!ASAP!! THANK YOUUUU!

mario62 [17]

Answer:

F = 9.675Hz

Explanation:

pls for certain reasons let us make

wavelength = $
frequency = F
V = velocity

3 loops : 6$/4 = L

6$/4 = 2

$ = 4/3 = 1.333

V = F x $

F = V/$

F = 12.9/1.333 = 9.675Hz

F = 9.675Hz

4 0

2 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy of 0.013 J.

Aleksandr [31]

answer

v = 4.2 $\frac{m}{s}$

set up an equation

we can use the formula for kinetic energy since we know mass and kinetic energy

K = $\frac{mv^{2}}{2}$

v = $\sqrt{\frac{2K}{m} }$

values

K = 0.013 J

m = 1.5g = 0.0015 kg

plug in values

v = $\sqrt{\frac{2*0.013}{0.0015} }$

v = 4.2 $\frac{m}{s}$

3 0

3 years ago