What is the period number in which helium is found?

Four toy racecars are racing along a circular race track. The cars start at the 3-o'clock position and travel CCW along the trac

xeze [42]

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

6 0

3 years ago

The Earth's Radius is 6.3710x106 m and mass is 5.9742x1024 kg. What is the acceleration due to gravity at Mount Everest (elevati

Shalnov [3]

Answer is

9.773m/s^2

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Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

-----------------------------------------------------------------------

hope this helps :)

3 0

3 years ago

Which of the following is an example of kinetic energy?

Elza [17]

The answer is C) a rolling bowling ball because kinetic energy is the energy of movement and potential energy is the energies of the others, since there are not in movement. i hope this helps.

7 0

3 years ago

Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field

swat32

Answer:

The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

$F= qE$

$mg=qE$

$V\times\rho\times g=qE$

$q=\dfrac{V\times\rho}{E}$

$q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}$

Put the value into the formula

$q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}$

$q=3.106\times10^{-16}\ C$

We need to calculate the quantization of charge

Using formula of quantization

$n = \dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}$

$n=1941.25$

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

7 0

3 years ago

Read 2 more answers

You throw a rock horizontally out of a 7th story window. You time that it takes 3.7 seconds to hit the ground, and measure that

Nady [450]

Since we are only looking at the vertical height, we can use the free fall equation to find the height:
h = 0.5*g*t^2, where h is height in m, g is acceleration due to gravity (9.81 m/s^2), and t is time in seconds
h = 0.5*(9.81 m/s^2)*(3.7 s)^2
h = 67.15 m
Therefore, the 7th floor window is 67.15 m above ground level.

5 0

3 years ago