Ask question

Ask question

All categories

3 years ago

15

A man can swim with a speed of 5m/s in calm water. if this man swims crosses a specific river his speed is 3m/s. if he takes the

minimum time to cross the river find the speed of the flow of water in the river...
plz answer me with all the steps!
I'll give u points ..mark brainliest and follow u

1 answer:

Alex787 [66]3 years ago

4 0

Answer:

The speed of the flow of water in the river is 4 m/s.

Explanation:

Given that,

Speed of man = 5 m/s

If this man swims crosses a specific river his speed is 3 m/s.

If he takes the minimum time to cross the river

Let the speed of flow of water be $v_{r}$

We need to calculate the speed of the flow of water in the river

Using formula for velocity

$v_{rs}^2=v_{r}^2+v_{s}^2$

$v_{r}^2=v_{rs}^2-v_{s}^2$

Where, $v_{s}$ = velocity of swimmer

$v_{sr}$ = relative velocity

$v_{r}$ = velocity of river

Put the value into the formula

$v_{r}^2=5^2-3^2}$

$v_{r}=\sqrt{5^2-3^2}$

$v_{r}=4\ m/s$

Hence, The speed of the flow of water in the river is 4 m/s.

You might be interested in

Find the current flowing out of the battery.

klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

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

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0

3 years ago

A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3

PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

7 0

2 years ago

Future passive of( win)

Inessa05 [86]

Answer:

three point charge positioned one x-axis if the charge and corresponding positions are +32Mc x=0 +20Mc x=40cm - 60Mc x=60cm find force 32Mc

Explanation:

7 0

2 years ago

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-

nikdorinn [45]

Answer:

The answer is " $60.74^{\circ}$ ".

Explanation:

Cavity and benzene should be extended in equal quantities.

$\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\$

$\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\$

$\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\$

5 0

2 years ago

Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =

Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

$v = 3.5596 \frac{m}{s}$

Explanation:

The motion have the velocity that will make the block move using:

$F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}$

$m_{1} = 1.2 Kg$

$m_{2} = 2.8 Kg$

$r = 0,45 m$

μ $s= 0,54$

Resolving:

$\frac{m_{1}*v^{2} }{r} = us*m_{1}* g + m_{2} * g$

$v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }$

$v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}$

$v^{2} = 12.6714 \frac{m^{2} }{s^{2} }$

$v = 3.559 \frac{m}{s}$

4 0

3 years ago