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Ivan
3 years ago
13

Determine the work done by the constant force. The locomotive of a freight train pulls its cars with a constant force of 15 tons

a distance of one-quarter mile.
Physics
1 answer:
Digiron [165]3 years ago
5 0

Answer:

5.92×10⁷ J

Explanation:

We'll begin by converting 15 tons to Newton. This can be obtained as follow:

1 ton = 9806.65 N

Therefore,

15 ton = 15 ton × 9806.65 N / 1 ton

15 ton = 147099.75 N

Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:

100 mi = 160934 m

0.25 mi = 0.25 mi × 160934 / 100 mi

0.25 mi = 402.335 m

Finally, we shall determine the Workdone. This can be obtained as follow:

Force (F) = 147099.75 N

Distance (d) = 402.335 m

Workdone (Wd) =?

Wd = F × d

Wd = 147099.75 × 402.335

Wd = 5.92×10⁷ J

Thus, the Workdone is 5.92×10⁷ J

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A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both
kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

q2 = −9.00 nC

As we know,

The potential difference between the two shell's difference will be:

⇒  \Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]

           =K[\frac{q1}{R2}-\frac{q1}{R1} ]

On substituting the values, we get

           =(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]  

       Δ =450 \ volts

3 0
3 years ago
What happens to the speed of the particles if the size of the particle is increased
bearhunter [10]
The particle slows down.
6 0
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How is thermal energy transferred during convection?
Lerok [7]
Conduction and <span>convection it involves particles.</span>
5 0
3 years ago
Read 2 more answers
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
Ganezh [65]

Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

5 0
3 years ago
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