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insens350 [35]
3 years ago
11

Tectonic plates are large segments of the earth's crust that move slowly. suppose one such plate has an average speed of 4.8 cm

per year. (a) what distance does it move in 40 seconds at this speed?
Physics
1 answer:
user100 [1]3 years ago
6 0
<span>one year is 365, 1 day is 24 hours, 1 hour is 60 minutes, 60 minutes is 60 seconds, thus (365 * 24 * 60 * 60) = 31,536,000 one year is equal to 31,536,000 seconds. the plate has a speed of 4.8 cm every 31,536,000 seconds. lets find out how far it goes in 40 seconds. (4.8/31,536,000)*40 = 0.00000608828 The plate moves 0.00000608828 cm every 40 seconds</span>
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Lelechka [254]

Answer:

6

Explanation:

The function which is inverse to exponentiation is called logarithm.

It is of the form

log\ x=y\\\Rightarrow log_{10}\ x=y\\\Rightarrow 10^y=x

The 10 here is the base of the logarithm. The logarithm with base 10 is called common logarithm

log_{10}1000000=y\\\Rightarrow 10^y=1000000\\\Rightarrow 10^6=1000000\\\Rightarrow y=6

So, log(1,000,000) = 6

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3 years ago
After six half-lives, what percentage of a radioactive sample will remain?
tatiyna
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4 0
3 years ago
Read 2 more answers
Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t
storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

T=2\pi\sqrt{\frac{I}{K}}

I=mR^{2}

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}

As dimensions of new balance wheel are one-third of their original values

R_{new}=\frac{R}{3}

\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega

5 0
3 years ago
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

4 0
2 years ago
Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this
LenaWriter [7]

Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

5 0
3 years ago
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