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3 years ago

I need help with these questions :(see image )

Physics

1 answer:

lara31 [8.8K]3 years ago

7 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

Angle of projection = θ(theta) = $30^0$

Initial Velocity = $V_0$ = $2x10^3$

Acceleration due to gravity = g = 9.8 m/s^2

Vertical Velocity = $V_Y$ = ?

Horizontal Velocity = $V_X$ = ?

Range of the Shell = R = ?

Maximum Height = H = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

Vertical Velocity is given by,

$V_Y = V_0Cos$ θ

$V_Y = (2x10^3)xCos(30)$

$V_Y = (2x10)^3x(0.866)\\\\V_Y = 1732.05 \frac{m}{s}$

Horizontal Velocity is given by,

$V_X = V_0Sin$ θ

$V_X = (2x10^3)xSin(30)\\\\V_X = (2000)x(0.5)\\\\V_X = 1000\frac{m}{s}$

Range is given by,

R = $\frac{V_0^2}{g}$ Sin2θ

$R = \frac{(2x10^3)^2}{10} x Sin(60)\\\\R = \frac{4x10^6}{10} x 0.866\\\\R= (4x10^5) x 0.866\\\\R = 346410.16 m$

Horizontal Velocity is given by,

<h2> $H = \frac{V_0^2Sin^2(theta)}{2g}\\\\\\H= \frac{(2x10^3)^2xSin(30)^2}{2x10}\\\\\\H= \frac{(4x10^6)x(0.25)}{20} \\\\H = 50000 m$ _____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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3 years ago

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

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Answer:

20.62361 rad/s

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Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

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3 years ago

A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses

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Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

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I need help with these questions :(see image )​

I need help with these questions :(see image )