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Ludmilka [50]
3 years ago
12

You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0

m/s. Part A Use the work-energy principle to find the rock's speed just as it left the ground. Express your answer in meters per second. vground
Physics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

Vi = 24.14 m/s

Explanation:

If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

Loss in K.E of the Rock = Gain in P.E of the Rock

(1/2)(m)(Vi² - Vf²) = mgh

Vi² - Vf² = 2gh

Vi² = Vf² + 2gh

Vi = √(Vf² + 2gh)

where,

Vi = Rock's Speed as it left the ground = ?

Vf = Final Speed = 17 m/s

g = 9.8 m/s²

h = height of rock = 15 m

Therefore,

Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]

Vi = √583 m²/s²

<u>Vi = 24.14 m/s</u>

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The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m=11,100 kg is the mass of the telescope

M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
Which image shows an example of the strong nuclear force in action?
dusya [7]

i think this is the incomplete page that you are showing but the answer is

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<h2><u>B</u></h2>
6 0
3 years ago
Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

y=\frac{\lambda (m+\frac{1}{2})D}{d}

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light used

In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

m = 13 is the order of the minimum

y=8.57 cm = 0.0857 m is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
The velocity of a car changes from 20 m/s east to 5 m/s east in 5 seconds. What is the acceleration of the car?
Klio2033 [76]
acceleration =  \frac{ v_{2}- v_{1}  }{t} = \frac{5-20}{5} =-3m/s^{2}
3 0
3 years ago
Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0
padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

3 0
3 years ago
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