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Mandarinka [93]
3 years ago
6

5. A family of ducks is swimming in a pond at a speed of 3 m/s when a gust of wind hits them. By the time they reach the other s

ide of the lack, the ducks are moving at a speed of 7 m/s. How long did it take for them to cross the lake if they accererated at a rate of 1 m/s2
Physics
1 answer:
ella [17]3 years ago
6 0

Answer:

The time taken by the duck to cross the lake is, t= 4 s

Explanation:

Given data,

The initial speed of the ducks, u = 3 m/s

The final speed of the ducks, v = 7 m/s

The acceleration of the duck, a = 1 m/s²

The formula for the acceleration is,

                               a = (v - u) / t

∴                               t = (v - u) / a

Substituting the given values in the above equation,

                                t = (7 - 3) / 1

                                  = 4 s

Hence, the time taken by the duck to cross the lake is, t= 4 s

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So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

\boxed{\sf{\bold{W = F \times s}}}

If the Angle Effect on Work

\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

With the following condition:

  • W = work that done by object (J)
  • F = force that applied (N)
  • s = shift or distance (m)
  • \sf{\theta} = angle of elevation (°)

<h3>Solution</h3>

We know that :

  • F = force that applied = \sf{1.41 \times 10^4} N
  • s = shift or distance = 84.9 m
  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

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1 year ago
A balloon having an initial temperature of 17.8°C is heated so that the volume doubles while the pressure is kept fixed. What i
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Answer:

T = 308.6 ^0 C

Explanation:

Here by ideal gas equation we can say

PV = nRT

now we know that pressure is kept constant here

so we will have

V = \frac{nR}{P} T

since we know that number of moles and pressure is constant here

so we have

\frac{V_2}{V_1} = \frac{T_2}{T_1}

now we know that initial temperature is 17.8 degree C

and finally volume is doubled

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Water flows over a section of Niagara Falls at the rate of 1.4 × 106 kg/s and falls 49.8 m. How much power is generated by the f
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Answer:

Power= 6.84×10⁸ W

Explanation:

Given Data

Niagara falls at rate of=1.4×10⁶ kg/s

falls=49.8 m

To find

Power Generated

Solution

Regarding this problem

GPE (gravitational potential energy) declines each second is given from that you will  find much the kinetic energy of the falling water is increasing each second.

So power can be found by follow

Power= dE/dt = d/dt (mgh)

Power= gh dm/dt

Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m

Power= 6.84×10⁸ W  

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