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lana66690 [7]
3 years ago
14

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost

? b) What is the ball's speed after the bounce? c) Where did the energy go?
Physics
1 answer:
tangare [24]3 years ago
4 0
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
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Which wavelength of light is the best choice when attempting to quantitatively relate solution absorbance and concentration?.
ANEK [815]

The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq)

you should choose a wavelength with maximum absorbance. In this case, you are using the scattered light, not the absorbed light as your signal. So you should avoid wavelengths where there are absorption peaks.

<h3>What is wavelength ?</h3>

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6 0
2 years ago
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
stepladder [879]

Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

5 0
3 years ago
A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph
Marina86 [1]

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
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h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

8 0
3 years ago
Please I need help with these 2 questions. Thank you.
Lemur [1.5K]
First one is D and Second one is B
7 0
3 years ago
Calculate the temperature change when 1000J of heat is supplied to 100g of water.<br> Please explain
jasenka [17]

Answer:

It is sensible heat- the amount of heat absorbed by 1 kg of water when heated at a constant pressure from freezing point 0 degree Celsius to the temperature of formation of steam i.e. saturation temperature

So it is given as - mass× specific heat × rise in temperature

i.e. 4.2 × T

4.2 × (100–0)

So it is 420kj

If you ask how much quantity of heat is required to convert 1 kg of ice into vapour then you have to add latent heat of fusion that is 336 kj/kg and latent heat of vaporization 2257 kj/kg (these two process occur at constant temperature so need to add rise in tempeature)

So it will be

Q= 1×336 + 1× 4.18 ×100 + 1× 2257

Q = 3011 kj

Or 3.1 Mj

Hope you got this!!!!!!

5 0
2 years ago
Read 2 more answers
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