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Galina-37 [17]
4 years ago
14

What is the most important

Physics
1 answer:
Andrew [12]4 years ago
4 0

Answer:

5

because while reading you should focus on your subjet when your mind will divert then you might reach to failure

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A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
4 years ago
If the relative density of a liquid is 0.34 what is the density of the liquid​
Elina [12.6K]

Answer:

<h2>3338.98 kg/m³</h2>

Explanation:

The formula for calculating the relative density of a substance is expressed as

Relative density of a liquid = Density of the liquid /density of water

Given relative density of a liquid = 0.34

Density of water 997kg/m³

Substituting into the formula we have;

Density of the liquid = Relative density of a liquid * density of water

Density of the liquid = 0.34 * 997

Density of the liquid = 3338.98 kg/m³

7 0
3 years ago
-What do you think happened to make the Moon look the way it does?
goldenfox [79]

Answer:  The physics of evolution had made the moon like it is today....Please watch this video from you tube about the evolution of the moon.

Explanation:

https://youtu.be/UIKmSQqp8wY

5 0
3 years ago
The moon's mass is 7.34x10-kg and it is 3.8x10m away from earth. Calculate the gravitational force of attraction between earth a
Nana76 [90]
Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.

<span>The formula for force of attraction between any two bodies in the universe
F  =  GMm / r^2.      (Newton's Universal law of Gravitation).

G = Universal gravitational constant, G = 6.67 * 10 ^ -11  Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22  kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
   = 3.8 x 10^8 m.

(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).


So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
  Punch it all up in your calculator.
 
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.

F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!</span>




8 0
4 years ago
Read 2 more answers
Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to t
Bogdan [553]

Complete question:

Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit? Fringe C is the central maximum.

Check the image uploaded.

Answer:

The difference is 1100 nm

Explanation:

A bright fringe creates constructive interference, the wavelength is always in a multiple form.

At center fringe, the difference is (550 nm)(0) = 0

For the first maximum, the difference is (550 nm)(1) = 550 nm

For the second maximum, the difference = (550 nm)(2) = 1100 nm

Thus, for nth maxima, the difference is (550 nm)(n)

From the image uploaded, C is located on the second maximum, therefore the difference is given as  (550 nm)(2) = 1100 nm

Therefore, the dot on the screen in the center of fringe E to the left slit is 1100 nm

3 0
4 years ago
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