If you weighed 130 pounds on earth, you would weigh ____

Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

Learn more about Momentum here:

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4 0

1 year ago

A hypothetical planet has a mass of 1.66 times that of Earth, but the same radius. What is gravitiy near its surface?

baherus [9]

Answer: 16.22 m/s^2

Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so

((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2

6 0

3 years ago

Read 2 more answers

7. All electromagnetic waves travel at the same speed in empty space, the

Vilka [71]

Answer:

TRUE

Explanation:

8 0

2 years ago

Read 2 more answers

A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the

Anton [14]

Answer:

$1.176\times 10^{-3} N$

Upward

Explanation:

We are given that

Mass of scarp paper, $m=120 mg=120\times 10^{-6}kg$

1mg= $10^{-6} mkg$

Distance,d =8 mm= $8\times 10^{-3} m$

Magnitude of electric force = $F_E= w=mg$

Where $g=9.8 m/s^2$

Substitute the values

$F_E=120\times 10^{-6}\times 9.8$

$F_E=1.176\times 10^{-3} N$

Gravitational force act in downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

6 0

3 years ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago

If you weighed 130 pounds on earth, you would weigh _____pounds on the moon