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2 years ago

10

A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 105 meters calculate his spe

ed

2 answers:

Bad White [126]2 years ago

8 0

The formula should be (2πr)/T. r=radius T=time. if time is in seconds the formula would look like this
$\frac{2\pi \times 105}{60 \times 2}$

ankoles [38]2 years ago

5 0

Speed = (distance covered) / (time to cover the distance)

Distance covered = (π · diameter) = 210π meters

Time to cover the distance = 2 minutes or 120 seconds

Speed = (210π meters) / (120 seconds)

Speed = 5.5 meters/sec

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Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

Zanzabum

(a) $3.3\cdot 10^{-6} Pa$

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

$p=\frac{I}{c}$

where

I is the intensity of the wave

c is the speed of light

In this problem,

$I=1000 W/m^2$

and substituting $c=3\cdot 10^8 m/s$ , we find the radiation pressure

$p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa$

(b) $4.4\cdot 10^{-8} m/s^2$

Since we know the cross-sectional area of the laser beam:

$A=6.65\cdot 10^{-29}m^2$

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

$F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N$

And then, since we know the mass of the atom

$m=5.01\cdot 10^{-27}kg$

we can find the acceleration, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2$

6 0

2 years ago

The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?

n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

$Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100$

The measured value is 2.02 meters and the actual value is 2.00 meters.

$V_{measured}=2.02\\\\V_{true}=2.00$

$Percent Error=\frac{2.02-2.00}{2.00} *100$

First, evaluate the fraction. Subtract 2.00 from 2.02

$Percent Error=\frac{0.02}{2.00}*100$

Next, divide 0.02 by 2.00

$PercentError=0.01 *100$

Finally, multiply 0.01 and 100.

$Percent Error=1\\Percent Error= 1 \%$

The percent error is 1%.

6 0

2 years ago

The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

kifflom [539]

Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s

6 0

2 years ago

Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in 8 s. Scenario C: 200 J of work is done in 10 s. W

hodyreva [135]

Using the equation P = W/t to solve your problem .

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8 0

1 year ago

Read 2 more answers

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$

We know that $K_c$ is 33Mpa*m^{0.5} and our Safety factor is 2,

$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

$F_{allow}=\sigma_{allow}*L_c*W$

$F_{allow} = 208.15*250*4$

$F_{allow} = 208.15kN$

That is the maximum tensile load before a catastrophic failure.

4 0

1 year ago