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Dima020 [189]
3 years ago
5

A 10 m long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 30 m/s when a 68

kg skydiver drops out by releasing his grip on the glider. What is the magnitude of the glider's velocity just after the skydiver lets go?
Physics
1 answer:
timama [110]3 years ago
5 0

Answer:

 v = 30.39 m/s

Explanation:

given,

mass of glider,M = 680 Kg

mass of the skydiver, m = 68 Kg

horizontal velocity,V = 30 m/s

when skydiver releases, velocity,v = 30 m/s

velocity if the glider,v' =  ?

use the conservation of momentum

M V = m' v' +  m v

m' = 680-68 = 612 Kg

680 x 30 = 612 x v + 60 x 30

612 v = 18600

 v = 30.39 m/s

since the skydiver's speed will be the same as before release

The glider should continue to travel at 30.39 m/s since there are no external forces acting on it

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Answer:

kinetic and potential

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A material allows charges to move freely through it. Which statement is the
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The answer is D- it has a high resistance and is a conductor


hope this helps !
4 0
3 years ago
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A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the
kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}

Now for time:

\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}

Solving for t,

\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}

The total time the kangaroo takes in the air is 2.3s.

3 0
1 year ago
A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie
Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.  

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

For the composite engine, its efficiency is given by

e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.  

But the work done can be written as  

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.  

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel  

each other.

 e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

               = Q_C1(T_C/T')

(T1 is the intermediate temperature)  

But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 =  -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by  

e_12=1-Tc/Th

This is same as the original Carnot engine.  

7 0
3 years ago
A pendulum of mass 5.0 ???????? hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal fo
Rudiy27

Answer:

a. 37.75°

b. 6.21 m

Explanation:

a. The horizontal force acting on a pendulum bob is given as:

F = mgsinθ

where m = mass of bob

g = acceleration due to gravity

θ = angle string makes with the vertical or angle of displacement

Making θ subject of formula, we have:

θ = sin^{-1}(\frac{F}{mg} )

θ = sin^{-1} (\frac{30}{5*9.8})

θ = 37.75°

The maximum angle of displacement is 37.75°

b. Period of a pendulum is given as:

T = 2\pi\sqrt{ \frac{L}{g} }

where L = length of string

Therefore, making L subject of formula:

L = \frac{gT^2}{4\pi^2}

L = \frac{9.8 * 5^2}{4\pi^2} \\\\\\L = 6.21 m

The string holding the pendulum has to be 6.21 m long.

4 0
3 years ago
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