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3 years ago

What happens to light at the surface of the water that creates the upside down image of the impalas?

Physics

2 answers:

lara31 [8.8K]3 years ago

7 0

By light reflection

mariarad [96]3 years ago

4 0

B the light reflection

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Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1360 N. Assume that the play

Alona [7]

Answer:

7.59Ns

Explanation:

Given parameters:

Force = 1360N

Time of contact = 5.85 x 10⁻³s

Unknown:

Impulse = ?

Solution:

The impulse of the ball is given as:

Impulse = Force x time

Impulse = 1360 x 5.85 x 10⁻³ = 7.59Ns

4 0

3 years ago

A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth

Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

$W=mg$

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center

The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

$800 N= W=mg=m \frac{GM}{R^2}$ (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

$200 N = W' = mg' = m \frac{GM}{(R+h)^2}$ (2)

If we divide eq.(1) by eq.(2), we get

$\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}$

$4=\frac{(R+h)^2}{R^2}$

By solving the equation, we find:

$4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0$

which has two solutions:

$h=-3R$ --> negative solution, we can ignore it

$h=R$ --> this is our solution

Since the Earth's radius is $R=6.4\cdot 10^6 m$ , the person should be at $h=R=6.4\cdot 10^6 m=6400 km$ above Earth's surface.

5 0

4 years ago

Which of the following is an example of an object with kinetic energy?

Kaylis [27]

Answer:

A and C

Explanation:

Both have mass and are in motion

6 0

3 years ago

Read 2 more answers

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago

What percentage of Earth is covered with water? 97 70 3 3.5 20 69

Leni [432]

Answer:

70 percent

Explanation:

the answer is 70 percent

7 0

3 years ago

Read 2 more answers