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3 years ago

Convert 11 milliseconds into seconds

Physics

2 answers:

Alex17521 [72]3 years ago

6 0

Answer:

0.011

Explanation:

shusha [124]3 years ago

5 0

0.011

divide the time value by 1000

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What will happen to the absorption and/or emission spectral lines of an object moving away from Earth at high speed?

Rufina [12.5K]

Answer:

c. They will be shifted toward the red end of the spectrum.

Explanation:

When the absorption or emission lines of a moving object away from from the earth at very high speed then the spectral lines will be shifted toward the red end of the spectrum.This effect can be easily understand by Doppler effect..

So the our option c is correct.

c. They will be shifted toward the red end of the spectrum.

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3 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

A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

Fantom [35]

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

- q × E = m × w² × r ..................1

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}r}$ q = m × w² × r ............2

and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ = 1.672 × $10^{-27}$ × 11304000² × 0.0120

linear charge density = -9.495 × $10^{-34}$ C/m

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3 years ago

A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?

Aleks04 [339]

25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N

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3 years ago

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Suppose a car of mass m is moving at a constant speed v of

SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

$\arcsin(v^2/(gR))$

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

$w_r = mg\sin(\theta)$