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Arte-miy333 [17]
3 years ago
13

.-. spooooooooooooooooooooooooooooooooooooooooooooooooooopy ok

Physics
1 answer:
Whitepunk [10]3 years ago
6 0

Answer:

omggghshsbsnznxjdbdn

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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998
wlad13 [49]

Answer:

<em>1988.05 rad/s^2</em>

<em></em>

Explanation:

The angular speed of the optical disk ω = 998.0 rad/s

the time taken to come to rest t = 0.502 s

The magnitude of the average angular acceleration ∝ = ω/t

∝ = 998.0/0.502 = <em>1988.05 rad/s^2</em>

7 0
3 years ago
HELP ASAP <br><br> What happens when the mantle melts?
ankoles [38]

It produces oceanic crust,

3 0
3 years ago
Read 2 more answers
Will binoculars work properly if their prisms (assume n = 1.50) are immersed in water (nwater= 1.33)? assume that binoculars wor
Amanda [17]

The binoculars will not work properly.

Please refer to the diagram attached.

Binoculars work on the principle of total internal reflection. The incident ray from air enters the prism through on face normally and it strikes the opposite face at an angle of 45°. The ray undergoes Total internal reflection if its critical angle is less than the angle of incidence,which , in this case is 45°.

The critical anglei_{c} is related to the refractive index of the material of glass μ as shown below.

\mu =\frac{1}{sini_c}

For a refractive index equal to 1.5, the critical angle is found as shown below.

sini_c=\frac{1}{\mu} \\ =\frac{1}{1.5} \\ =0.666

Take the inverse of the value 0.666 to determine the critical angle.

sin^{-1} (0.666)=41.8^o

The critical angle in air is less than the angle of incidence, and hence total internal reflection occurs in air.

When the binoculars are immersed in water, the ray passes into the glass through water. therefore, the refractive index of the prism when immersed in water is given by,

\mu_g_w=\frac{\mu_g}{\mu_w}

Therefore the critical angle <em>c</em> in this case is given by,

sinc=\frac{\mu_w}{\mu_g}

Substitute 1.33 for \mu_w and 1.5 for \mu_g.

sinc=\frac{\mu_w}{\mu_g} \\ =\frac{1.33}{1.5} \\ =0.8866

Take the inverse of the value 0.8866.

c=sin^{-1} (0.8866)\\ =62^o

Since the critical angle of the prism when immersed in water, is 62°, which is greater than the angle of incidence of 45° required for viewing the object, the binoculars which are set for Total reflection in air, will not function when immersed in water.

4 0
4 years ago
To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th
il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).<span>

The amount of weight displaced or the amount of buoyant force is: </span>

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

 

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

<span>average density = 1.07 kg/L = 1.07 g/mL</span>

5 0
3 years ago
Read 2 more answers
What is number 30,31 in this picture?
Maslowich
On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C

Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)

on the DT graph im going to assume that movement B is a positive velocity with A and C being negative. 
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity. 

Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question. 
i'll skip my working out as its the same as before:

C, B, A and then D

the same idea as on Q30 applies to Q31 part b, 

D,C,B then A 
6 0
3 years ago
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