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3 years ago

A person paddles down river at an average speed of 4km

Physics

2 answers:

adoni [48]3 years ago

7 0

Nice paddling. Thanks for sharing.
Do you have some question to ask ?

almond37 [142]3 years ago

7 0

Is that a question or are you just telling us

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of a

alukav5142 [94]

Answer:

# of Snickers bars 2

Explanation:

Power output= 0.30 HP

=0.3*746

= 0.30 HP (746 W=1.00 HP)

= 224 W

time required 2 h 49 m = 10140 seconds

Since power is work divided by time, then work is:

Work done by the jet = P*t

= 224 *(10140)

= 2.3 MJ (2.3 x $10^{6}$ J)

Converting MJ to Cal

2.3 MJ=549 Cal

# of Snickers bars = 549 Cal / 280 Cal

= 2.0 bars (rounded from 1.96)

8 0

3 years ago

Please list the following for the Element, HYDROGEN

bagirrra123 [75]

Answer:

1 Proton, 1 Electron, No Neutrons

Group 1, Period 1

Gases

3 0

2 years ago

ILI

Nikolay [14]

The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

$\text {Kinetic energy}=\frac{1}{2} m v^{2}$

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.

The final kinetic energy is

$\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5$ = 7500 J

As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.

Work done = Final Kinetic energy - Initial Kinetic energy

Thus the work utilized for moving the car is

Work done = 7500 J - 0 J = 7500 J

Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

7 0

3 years ago