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omeli [17]
3 years ago
9

My body is not just something that I have, my body is also I am

Physics
2 answers:
sertanlavr [38]3 years ago
8 0

Answer:

thats pretty cool k byeeeeeeee

Bond [772]3 years ago
4 0

Answer:

um ok

Explanation:

i guess

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What are 3 states of matter.write 1 fact for each state of matter
Masteriza [31]
The three states of matter are:
Solid- The atoms are packed tightly.
Liquid-Liquids have a Definite volume but not a definite shape.
Gas-No definite volume and no definite shape;
7 0
3 years ago
Read 2 more answers
A ball with mass of 0.050 kg is dropped from a height of h1 = 1 .5 m. It collides with the floor, then bounces up to a height of
Iteru [2.4K]

Answer:

Explanation:

Impulse of reaction force of floor = change in momentum

Velocity of impact = √ 2gh₁

= √ 2 x 9.8 x 1.5 = 5.4 m /s.

velocity of rebound = √2gh₂

= √ 2x 9.8 x 1

= 4.427 m / s.

Initial momentum = .050 x 5.4 = .27 kg m/s

Final momentum = .05 x 4.427 = .22 kg.m/s

change in momentum = .27 - .22 = .05 kg m/s

Impulse = .05 kg m /s

Impulse = force x time

force = impulse / time

.05 / .015 = 3.33 N.

kinetic energy = 1/2 m v²

Initial kinetic energy = 1/2 x .05 x 5.4²

= 0.729 J

Final Kinetic Energy =1/2 x .05 x 4.427²

= 0.489 J

Change in Kinetic energy =0 .24 J

Lost kinetic energy is due to conversion of energy into sound light etc.

4 0
4 years ago
High energy waves have short wave lengths, true or false?
Anna71 [15]
True, the wavelength dies down due to high frequency and low amptitude. 
5 0
3 years ago
A car of mass m=1000kg is traveling at speed v and brakes. The skid marks are 20m long and the coefficient of kinetic friction i
scoray [572]

Answer:

v = 14 m/s

  = 31.3 mph

The answer would be the same if the mass of the car were 2000 kg

Explanation:

Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.

V^2 = v^2 + 2×a×Δx

Now V, the final velocity is zero as the car stops

0 = v^2 + 2×a×Δx

v^2 = -2×a×Δx

v =√-2×a×Δx    .....*

Now applying Newton's Second Law

Fnet = m×a

-Fk = m×a

-μ×N = m×a

-μ×m×g = m×a (The mass cancels out)

a =  -μ×g

Substituting the value of a back to equation *

v = √-2×(-μ×g)×Δx  

v = √-2×(-0.5×9.8)×20

v = 14 m/s

Therefore the speed the car was travelling with v = 14 m/s

which is equal to 31.3 mph

Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.

8 0
3 years ago
Consider a single slit diffraction pattern for awavelength of 589 nm porjected ona screen that is 100m from a slit of width 0.25
3241004551 [841]

Answer:

c

Explanation:

6 0
3 years ago
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