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3 years ago

7

2.A girl of mass 50kg climbs a flight of a stair having 40steps each of 30m high in 10 s. Calculate the power developed by the g

irl.

1 answer:

nikdorinn [45]3 years ago

8 0

Answer:

${ \boxed{ \pmb{power = \frac{force \times distance}{time} }}} \\$

force → (50 × 10) = 500 N
distance → (30 × 40) = 1200 m
time → 10 seconds

${ \rm{power = \frac{500 \times 1200}{10} }} \\ \\ { \rm{power = \frac{600000}{10} }} \\ \\{ \underline{ \underline{ \rm{ \dashrightarrow \: \: power = 60,000 \: newtons \: \: }}}}$

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A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert

Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

$R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s$

To find:

Max Height $h$ achieved

Calculation:

1) Using the $3^{rd}$ equation of motion, we know

$2*a*s = Vf^{2} - Vi^{2}$

2) In terms of gravity $g$ height $h$ and the vertical component of Velocity $Vf , Vi$ .

3) As $Vf = 0$ as at maximum height the vertical component of velocity is zero maximum height achieved

$2*g*h = Vf^{2} -Vi^{2}$

putting values

4) $h = 0.784 m/s$

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0

4 years ago

A small racquet ball launcher is set up 5 meters from a 12 meter tall wall. It launches a racket ball at 30 m/s at an angle of 4

Strike441 [17]

Answer:20.82 m

Explanation:

Given

distance between wall and launcher=5 m

initial velocity=30 m/s

Launching angle $=45^{\circ}$

Height of wall=12 m

maximum height by ball

$h_{max}=\frac{u^2sin^2\theta }{2g}$

$h_{max}=\frac{30^2sin^{2}45}{2\times 9.8}$

h=22.95 m

$y=xtan\theta -\frac{gx^2}{2u^2cos^\theta }$

y=4.72 m

so ball will strike with wall

and for perfect restitution final velocity of ball will be same as the horizontal velocity before its impact, only direction will be opposite and vertical velocity will be zero

thus it seems as if someone throw the ball with horizontal velocity of 30cos45 from a height of 4.72

Time required to cover 4.72 m

$t=\sqrt{\frac{2h}{g}}$

$t=\sqrt{\frac{9.45}{9.8}}$

t=0.981 s

Horizontal distance traveled in this time

$R=ucos45\times t$

$R=30\times cos45\times 0.981$

R=20.82 m

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3 years ago

A ball is thrown straight up into the air with a speed of 10 m/s . If the ball has a mass of 0.3 kg how high does the ball go ?

Damm [24]

<h3><u>Answer;</u></h3>

= 5.102 m

<h3><u>Explanation;</u></h3>

The total energy, i.e. sum of kinetic and potential energy, is constant.

That is; E = KE + PE

Initially, PE = 0 and KE = 1/2 mv^2

At maximum height, velocity=0, thus, KE = 0 and PE = mgh

Since, total energy is constant (KE converts to PE when the ball is rising).

Therefore, KE = PE

or, 1/2 mv^2 = mgh

or, h = v^2 /2g

= 10^2 / (2x9.8)

=<u> 5.102 m</u>

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3 years ago

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A jetliner is moving at a speed of 245m/s. The vertical component of the plane's velocity is 40.6 m/s. Determine the magnitude o

Harman [31]

Answer:

Horizontal component of velocity = 241.61 m/s

Explanation:

Given that,

Velocity of a jetliner is 245 m/s

The vertical component of the plane's velocity is 40.6 m/s, $v_y=40.6\ m/s$

We need to find the magnitude of the horizontal component of the plane velocity.

The resultant velocity is given by :

$v=\sqrt{v_x^2+v_y^2}$

Where, $v_x$ is the horizontal component of the plane velocity

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So, the magnitude of the horizontal component of the plane velocity is 241.61 m/s.

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3 years ago

a 5.00kg crate is on a 21.0° hill. Using X-Y axes tilted down the plane, what is the x-component of the weight?

Elodia [21]

M g = 5.00 (9.81) = 49.05

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4 years ago

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