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Contact [7]
3 years ago
7

You own a yacht which is 14.5 meters long. It is motoring down a canal at 10.6 m/s. Its bow (the front of the boat) is just abou

t to begin passing underneath a bridge that is 30.0 m across. How much time is required until its stern (the end of the boat) is no longer under the bridge?
Physics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

4.198 seconds

Explanation:

The total length of the yacht is 30m and the width of the bridge is 14.5m.

The speed of the yacht is 10.6m/s.

We can assume the current of the canal has a speed of 0m/s.

To calculate the answer we will break it into two parts

1) The time it takes for the bow to travel from the start of the bridge to the end

<em>Formula Speed = Distance / Time</em>

Speed = 10.6m/s

Distance = 30m

Time = T1

T1 = 2.8301s

2) The time it takes for the stern to cross the end of the bridge after the bow

<em />

<em>Formula Speed = Distance / Time</em>

<em />

Speed = 10.6m/s

Distance = 14.5m

Time = T2

T2 = 1.3679s

3) The Total Time is

Total = T1 + T2

Total = 4.198s

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
3 years ago
Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. What is the h
Viktor [21]
The velocities and the speed build a triangle, where the 1.7 m/s are the hypotenuse and the x-velocity and y-velocity are the other sides. 

<span>So the x-velocity is: speed*cos(angle) </span>

<span>now plug in </span>
<span>x=1.7 m/s * cos(18.5)=1.597 m/s </span>


3 0
3 years ago
Which portion of the electromagnetic spectrum is used in a microscope?
erik [133]
Answer is C.
 
Optical microscope involves passing visible light transmitted through or reflected from the sample through a single or multiple lenses to allow a magnified view of the sample.

8 0
3 years ago
[100 POINTS] Which of the following do not make their own energy through nuclear fusion? Select all that apply.
Ksenya-84 [330]

Answer:

The correct answer is :

A . Giant star

B . proto star

F .main sequence star

Explanation:

hope this helps

7 0
3 years ago
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