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3 years ago

Which example involves the transformation of chemical energy directly into light energy?

Physics

2 answers:

solong [7]3 years ago

6 0

Answer:

Explanation:

A glow stick involves the transformation of chemical energy into light energy. The chemical energy is stored.

aleksley [76]3 years ago

4 0

Yes, you're right, it's a glow stick!

a glow stick has two substances in it, which are separated from each other and breaking of the glow stick makes them combine and release energy in the form of a light. This process is also known as chemiluminescence. They don't require any other energy sources!

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2 Points

yulyashka [42]

The force applied to lift the crate is 171 N

Explanation:

The lever works on the principle of equilibrium of moments, so we can write:

$F_i d_i = F_o d_o$

where

$F_i$ is the force in input

$d_i$ is the arm of the input force

$F_o$ is the output force

$d_o$ is the arm of the output force

For the lever in this problem, we have:

$d_i = 0.25 m$

$d_o = 0.19 m$

$F_i = 130 N$ (force applied)

Solving the equation for $F_o$ , we find the force applied to lift the crate:

$F_o = \frac{F_i d_i}{d_o}=\frac{(130)(0.25)}{0.19}=171 N$

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

6 0

2 years ago

What is the mass of a stone moving at a speed of 15 m/s and having a monument at 7.1 kg meters per second

adell [148]

Answer:

<h3>The answer is 0.47 kg</h3>

Explanation:

The mass of the object given it's momentum and velocity can be found by using the formula

$m = \frac{p}{v} \\$

where

p is the momentum

v is the velocity

We have

$m = \frac{7.1}{15} \\ = 0.4733333...$

We have the final answer as

Hope this helps you

4 0

2 years ago

Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve

dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

8 0

2 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0

2 years ago

A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w

Rina8888 [55]

<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span> <span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span> <span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)=<span> <span>1960.74 rev/min
</span></span>

5 0

3 years ago