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Paladinen [302]

3 years ago

12

It took a crew 3 h 20 min to row 5 km upstream and back again. if the rate of flow of the stream was 2 km/h, what was the rowing

speed of the crew in still water?

1 answer:

Basile [38]3 years ago

8 0

Suppose rowing speed is V km/s.

So, upstream speed = V-2

Downstream speed = V+ 2

Total time is taken 3 h 20 min to cover up and down journey of 5 km.

So, total time = upstream time + downstream time

$3. 3 \ hours = \frac{5 km}{ V- 2} + \frac{5 km}{V+2} \\\\\ 3.3 (V^2-2^2) = 10V\\\\V^2- 4-3V= 0$

$V^2 -4V+V -4=0 \\\\\ V( V-4) +1 (V-4)=0$

$V= 4 km/h$ and $V= -1 \ km/h$ , we take only positive velocity.

Thus, the the rowing speed of the crew in still water is 1 km/s.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

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4 years ago

A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force

MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = $\frac{D}{2}$

= $\frac{200}{2}\ m$

= 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

Force F = $\frac{mv^2}{R}$

= $\frac{1500\times (25)^2}{100}\ N$

= 9375 N. [ANS}

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professor190 [17]

When the car moves and makes a sound that is louder that when the car is just sitting there

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A car travelling at 15m/s comes to a rest in a distance of 14m when the brakes are applied. Calculate the deceleration of the ca

vlada-n [284]

Answer:

force of the breaks is 6650 N, direction opposite to direction of movement

Explanation:

6 0

4 years ago

Read 2 more answers

Science help answer all

wlad13 [49]

Idk but i hope you figure it out :)

4 0

3 years ago