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ElenaW [278]
3 years ago
11

The strength of the force depends on which two factors?

Physics
1 answer:
qwelly [4]3 years ago
7 0

Answer:

The answer is....

Explanation:

How hard the surfaces push together and the types of surfaces involved.

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The standard unit of brightness is called the candela.<br> True<br> False
Westkost [7]

Answer:

TRUE

Explanation:

3 0
3 years ago
Would the tide be higher when the moon is on the same side of earth as New Brunswick or on the opposite side why
Sindrei [870]


Tide is the alternating pattern of rising and falling sea level with respect to land.

There are three types of tide relative to the position of the moon and sun to that of the earth.

1. Spring tides. Occur when the moon sun and earth arrange themselves more or less in a straight line, like an arrow. These tides are the highest and lowest tides.

2. Perigean spring tides. Occur when the new moon or full moon closely aligns with perigee, the closest point  to the earth in the moon's orbit. These tides are not as high or low as the spring tides. 

3. Neap tides. Occur when the sun and the moon are at right angles as seen from earth.The  tides are at minimum range.

3 0
2 years ago
Calculate the speed of light if the wavelength is 3cm and the period is 1*10^-10
ioda
First you find the frequency which is. 1/T(period) so u get 1*10^10
Now that you know frequency u can solve for speed: s= wavelength* frequency, but don't forget to convert 3cm into meters: 3/100= .03m so now you do (1*10^10)(.3)= 300,000,000 m/s or 3*10^8 m/s

Hope this helped :))
6 0
3 years ago
During takeoff, an airplane goes from 0 to 56 m/s in 9 s. How fast is it going after 4 s?
meriva
24 miles per second
56/9=6
6+6+6+6=24
8 0
3 years ago
A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe
Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

So the total moment of inertia at the beginning is

I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

where I_2 is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

r=\frac{0.450 m}{2}=0.225 m

so the moment of inertia of the rings is

I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

7 0
3 years ago
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