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2 years ago

14

Taylor drives 5 miles in 10 minutes. She stops at a light for 2 minutes. She then travels another 10 miles in 8 minutes. What wa

s her average speed?

1 answer:

attashe74 [19]2 years ago

5 0

45 miles per hour is the answer to the probably.

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The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps

kogti [31]

Answer:

$2.82942\times 10^{24}\ W\m^2$

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = $\frac{d}{2}=\frac{30}{2}=15\ mu m$

P = Power of the laser = $2\times 10^{15}\ W$

A = Area = $\pi r^2$

Intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2$

The light intensity within this spot is $2.82942\times 10^{24}\ W/m^2$

8 0

3 years ago

An atom containing 16 protons, 31 neutrons, and 16 electrons would have an atomic mass of

Eddi Din [679]

<span>47 amu is the correct answer</span>

7 0

3 years ago

Read 2 more answers

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Nimfa-mama [501]

Answer:

25 m/s

Explanation:

Given that:

Initial speed, u = 4 m/s

Final velocity, V = 11 m/s

Time, t = 8 seconds

t2, = 16 seconds

Acceleration, a= (change in velocity) / time interval

a = (11 - 4) / 8

a = 7 / 8 = 0.875m/s²

Final velocity, v2 ;

Acceleration * t2

0.875 * 16 = 14

V2 = 14 m/s

Final speed : v + v2 = (11 + 14)m/s = 25m/s

3 0

3 years ago

A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th

Alex73 [517]

Answer:

$V_f = 13.8863 \angle 60.89\°$

Explanation:

Our values are,

$m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s$

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

$m_1v_1=(m_1+m_2)v_fcos\theta$

For the particular case on the Y axis, we do the same with the speed of object 1.

$m_2v_2=(m_1+m_2)v_fsin\theta$

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

$Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°$

Replacing in any of the two functions, given above, we will find the final speed after the collision,

$(950)(16)=(950+1300)V_fcos(60.89)$

$V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}$

$V_f = 13.8863 \angle 60.89\°$

8 0

3 years ago

if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a

Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

4 0

2 years ago