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netineya [11]
3 years ago
9

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi

ne. When he recovers, he sees that he is fast overtaking a camper ahead in his lane moving at the slower speed of 15.1 m/s. He hits the brakes as fast as he can (his reaction time is 0.31 s). If he can decelerate at 2.5 m/s2, what is the minimum distance between the driver and the camper when he first sees it so that they do not collide
Physics
1 answer:
valkas [14]3 years ago
5 0

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

x=vt=(15.1m/s)(13.49s)=203.69m   (1)

Next you calculate the distance that teacher has traveled for 13.6s:

x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m  (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

x_{min}=692.88m-203.69m=489.19m

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The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.
ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

\lambda=\frac{h}{p}

where

h=6.6 \cdot 10^{-34} Js is the Planck constant

p is the momentum of the particle


In this problem, the momentum of the electron is equal to the product between its mass and its speed:

p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m


So, the answer is True.

7 0
3 years ago
HELPPP PLEASE !!!!!!!
vlabodo [156]

Explanation:

a) d = ½.a.t²

200 = ½(4)t²

200 = 2t²

t² = 200/2

t² = 100

t =√100 = 10 s

b) Vt = a. t

= 4(10)

= 40 m/s

c) V av. = d/t = 200/10 = 20m/s

6 0
2 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed v of the geosynchronous satellite is given by:

v=\frac{2\pi R}{T}

Because 2\pi R is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

F_c=\frac{mv^{2} }{R}

If we substitute the expression for v in this formula, we get:

F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}

Since the centripetal force is the gravitational force F_g between the satellite and the Earth, we know that:

F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }

Where G is the gravitational constant (G=6.67*10^{-11} Nm^{2}/kg^{2}) and M is the mass of the Earth (M=5.97*10^{24}kg). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0
3 years ago
Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp
ankoles [38]

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

4 0
3 years ago
I need help on the data section of the circuit design lab on Edg.
Arte-miy333 [17]

I hope it's not too late, but here you go

8 0
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