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2 years ago

A moving ball with a momentum of 25 kg m / s collides head-on with a wall.

Physics

2 answers:

Harman [31]2 years ago

4 0

Answer: D. 1000N

Explanation:

p1 = 25 kgm/s, p2 = -25 kgm/s, t = 50 ms = 50 * 10^(-3) s = 0.05 s

change in momentum = initial momentum – final momentum

Mathematically represented as:

Δp = p1 - p2 = 25 - (-25) = 50 kgm/s

Δp = 50 kgm/s

change in momentum = Force * time interval

Mathematically represented thus:

Δp = F * Δt

Making F the subject of the equation, we have

F = Δp/Δt

F = 50/0.05

F = 1000 N

Yanka [14]2 years ago

3 0

Answer:

d 1000N

Explanation:

change in momentum = 25 -(-25) = 50

by impuls momentum theorm

F.Δt = p2 - p1

f = p2 - p1 / Δt

f = 50/50*10^-3

so f = 1000N

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An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (<u>becarefull, gravity at this point is no longer</u> $9.8m/s^{2}$ <u>because we are not in the surface anymore</u>)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

$mg=200.4 N$ .

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2 years ago

Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up

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2 years ago

A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen

saw5 [17]

Answer:

The electric field strength is $4.5\times 10^{4} N/C$

Solution:

As per the question:

Area of the electrode, $A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}$

Charge, q = 50 nC = $50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m$

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

$\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}$

Now, the electric field strength of the electrode is:

$\vec{E} = \frac{\sigma}{2\epsilon_{o}}$

where

$\epsilon_{o} = 8.85\times 10^{- 12} F/m$

$\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}$

$\vec{E} = 4.5\times 10^{4} N/C$

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2 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

Greeley [361]

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2 years ago

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A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of

son4ous [18]

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

7 0

2 years ago