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Alex17521 [72]
3 years ago
15

A boat is able to move throught still water at 20m/s. It makes a round trip to a town 3.0km upstrea. If the river flows at 5m/s,

the time required for this round trip is
Physics
1 answer:
Kryger [21]3 years ago
8 0

Answer:

t=320s

Explanation:

Given Data

Boat speed=20 m/s

River flows=5 m/s

Total trip of distance d=3.0km = 3000m

To find

Total time taken

Solution

As

Velocity=distance/time\\time=distance/velocity\\

Here we have two conditions

First when boat moves upward and the river pushing back.then velocity is given as

velocity=20m/s-5m/s

velocity=15 m/s

Time for that velocity

t_{1} =distance/velocity\\t_{1}=\frac{3000m}{15m/s}\\ t_{1}=200s

Now for second condition when river flows and boat speed on same direction

velocity=20m/s+5m/s

velocity=25 m/s

Time taken for that velocity

t_{2}=distance/velocity\\t_{2}=\frac{3000m}{25m/s}\\ t_{2}=120m/s

Now the total time

t=t_{1}+t_{2}\\t=(200+120)s\\t=320s

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A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of
Naya [18.7K]

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t²    ..............1

and

depth = velocity of sound × time    .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1  and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

3 0
3 years ago
A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a diff. One friend says to the other
olga55 [171]

Answer:

The statement is not correct.

Explanation:

To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.

Data obtained from the question include:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 3 s

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 3)

v = 0 + 29.4

v = 29.4 m/s

Thus, the velocity of the car after 3 s is 29.4 m/s.

Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.

8 0
3 years ago
"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of
Alja [10]

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

6 0
3 years ago
Read 2 more answers
The results of a dart game were precise but not accurate.The accepted value of the game was the center of the dartboard.Which co
Vlada [557]

Answer:

The darts hit the same general area of the board.

8 0
3 years ago
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Coherent light with wavelength 610 nm passes through two very narrow slits, and theinterference pattern is observed on a screen
andriy [413]

Answer:

= 1220 nm

= 1.22 μm

Explanation:

given data:

wavelength \lambda = 610 nm = 610\times  10 ^{-9} m

distance of screen from slits D = 3 m

1st order bright fringe is 4.84 mm

condition for 1 st bright is

d sin \theta =\lambda     ---( 1)

andtan \theta = \frac{y}{ D}

\theta = tan^{-1}\frac{(y }{D})

= 0.0924 degrees

plug theta value in equation 1 we get

d sin ( 0.0924) = 610 \times 10 ^{-9}

d = 3.78\times 10^{-4} m

condition for 1 st dark fringe

d sin \theta =\frac{λ'}{2}

\lambda '= 2 d sin\theta

= 2λ    since from eq (1)

= 1220 nm

= 1.22 μm

7 0
3 years ago
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