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Art [367]
2 years ago
7

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

throwing velocity?
Physics
1 answer:
DerKrebs [107]2 years ago
7 0

The ball's horizontal and vertical velocities at time t are

v_x=v_{xi}

v_y=v_{yi}-gt

but the ball is thrown horizontally, so v_{yi}=0. Its horizontal and vertical positions at time t are

x=v_{xi}t

y=20.0\,\mathrm m-\dfrac g2t^2

The ball travels 22 m horizontally from where it was thrown, so

22\,\mathrm m=v_{xi}t

from which we find the time it takes for the ball to land on the ground is

t=\dfrac{22\,\rm m}{v_{xi}}

When it lands, y=0 and

0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2

\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}

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A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th
NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

PE= \frac{GMm}{d}

The kinetic energy can be written as,

KE= \frac{1}{2} mv^2

Where,

G = 6.67*10^{-11}m^3/kg\cdot s^2Gravitational Universal Constant

m = 5.972*10^{24}kg Mass of Earth

h = 56*10^6m  Height

r = 6.378*10^6m Radius of Earth

From the conservation of energy:

\frac{1}{2} mv^2 = \frac{GMm}{d}

Rearranging to find the velocity,

v = \sqrt{\frac{2Gm}{d}} \rightarrow  Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

d = r+h

v = \sqrt{\frac{2Gm}{r+h}}

Replacing the values we have that

v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}

v = 3.6km/s

Therefore the escape velocity is 3.6km/s

3 0
2 years ago
The gravitational force between two asteroids is 6.2 × 108 n. asteroid y has three times the mass of asteroid z. if the distance
SOVA2 [1]
Let m =  mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.

Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)

The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
     = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
    = 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)

Answer: 1.1x10¹⁶ kg
4 0
2 years ago
Read 2 more answers
A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
Jet001 [13]

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0
2 years ago
Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem
Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

V= 5.5\times 6.5\times 3\ m^3

V=107.25\ m^3

V=107.25\times 1000 L

V= 107250 L

At STP number of moles given as

n=\dfrac{V}{V_{at\ S.T.P.}}

V=22.4 L at S.T.P.

n=\dfrac{107250}{22.4}\ moles

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

4 0
2 years ago
Read 2 more answers
A 5.0-n projectile leaves the ground with a kinetic energy of 220 j. at the highest point in its trajectory, its kinetic energy
NikAS [45]
First, we get the difference between the kinetic energies such that,
             difference = (220J - 120J)
             difference = 100 J
The difference in kinetic energy is the equivalent of the potential energy which is calculated through the equation,
              PE = mgh
To calculate for the height, we derive the equation in a form,
           h = PE/mg
The product of the mass and acceleration due to gravity is the weight. 
                   h = (100 J) / (5 N)
                   h = 20 m

<em>Hence, the answer is 20 m. </em>
3 0
2 years ago
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