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2 years ago

An object is released from height of 17m. The object will hit the ground approximately in

1 answer:

5 0

$\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}$

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A satellite in earth orbit has a mass of 99 kg and is at an altitude of 2.02 106 m. (assume that u = 0 as r â â.) (a) what is th

Ierofanga [76]

(a) What is the potential energy: PE = -G * M * m/r

Where: M is the mass of the earth which is 5.98 * 10^24 kg.

m is the mass of the satellite.

r is the space from the center of the earth to the satellite

To conclude this distance add the radius of the earth to the altitude. Radius of the earth is 6.38 * 10^6 meters.

r = 6.38 * 10^6 + 2.02 * 10^6 = 8.38 * 10^6

PE = 6.67 * 10^-11 * 5.98 * 10^24 * 99/8.38 * 10^6 = 4.71240095 * 10^9 J

(b) magnitude of the gravitational force exerted by the Earth

Fg = G * M * m/r^2

Fg = 6.67 * 10^-11 * 5.98 * 10^24 * 99/(8.38 * 10^6)^2 = 562.3078873 N

8 0

3 years ago

A rescue airplane is diving at an angle of 37º below the horizontal with a speed of 250 m/s. It releases a survival package when

alisha [4.7K]

Answer:

<em>The correct option is 1. 720 m</em>

Explanation:

<u>Projectile Motion</u>

When an object is launched in free air (no friction) with an initial speed vo at an angle $\theta$ , it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.

$\displaystyle V_o=250\ m/s$

$\displaystyle \theta =-37^o$

The initial velocity has these components in the x and y coordinates respectively:

$\displaystyle V_{ox}=250\ cos(-37^o)=199.7\ m/s$

$\displaystyle V_{oy}=250\ sin(-37^o)=-150.5\ m/s$

And we know the plane has an altitude of 600 m, so the package will reach ground level when:

$\displaystyle y=-600\ m$

The vertical distance traveled is given by:

$\displaystyle y=V_{oy}\ t-\frac{g\ t^2}{2}=-600$

We'll set up an equation to find the time when the package lands

$\displaystyle -150.5t-4.9\ t^2=-600$

$\displaystyle -4.9\ t^2-150.5\ t+600=0$

Solving for t, we find only one positive solution:

$\displaystyle t=3.6\ sec$

The horizontal distance is:

$\displaystyle x=V_{ox}.t=199.7\times3.6=720\ m$

The correct option is 1. 720 m

6 0

3 years ago

A measurement of a change<br> in distance over time

aleksandrvk [35]

Explanation:

<h2>Speed is the rate of change of distance with time.</h2>

7 0

3 years ago

A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the m

miskamm [114]

Answer:

4.9x10^-6T

Explanation:

See attached file

6 0

3 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago