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mariarad [96]
3 years ago
12

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t

he roof? Note: the acceleration of an object due to earth's gravity is 32 f t sec 2
Physics
1 answer:
Neko [114]3 years ago
6 0

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

v_f^2 - v_i^2 = 2 a d

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

v_i = 0

also the acceleration of the stone is due to gravity which is given as

g = 32 ft/s^2

now we have

v_f = 180 ft/s

so from above equation

180^2 - 0 = 2(32)d

d = 506.25 ft

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Define wheel and axle in science terms
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Answer:

A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.

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6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord
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Answer:

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4 0
3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
2 years ago
A skydiver has jumped out of a plane and is falling faster and faster. what forces are present in this situation
kompoz [17]
Gravity and air resistance 

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5 0
3 years ago
An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e
frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

Tsin\theta= \frac{mv^2}{r}

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

Tcos\theta = mg

Matching both expressions with respect to the tension we will have to

T = \frac{\frac{mv^2}{r}}{sin\theta}

T = \frac{mg}{cos\theta}

Then we have that,

\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

\frac{mv^2}{r} = mg tan\theta

Rearranging to find the velocity we have that

v = \sqrt{grTan\theta}

The value of the angle is 14.5°, the acceleration  (g) is 9.8m/s^2 and the radius is

r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}

r = \frac{11.1}{2}+2.41

r = 7.96m

Replacing we have that

v = \sqrt{(9.8)(7.96)tan(14.5\°)}

v = 4.492m/s

Therefore the speed of each seat is 4.492m/s

6 0
3 years ago
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