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photoshop1234 [79]
3 years ago
14

A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T

he ball lands on the ground 2.9 s after it is launched. What is the height of the building (in m)?
Physics
1 answer:
Mamont248 [21]3 years ago
3 0

Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

\implies y_0\approx16.258\,\mathrm m

so the building is about 16 m tall (keeping track of significant digits).

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In 1990, Dave Campos of the United States rode a special motorcycle called the Easyrider at an average speed of 518 km/h. Suppos
maks197457 [2]

The distance travelled during the given time can be found out by using the equations of motion.

The distance traveled during the time interval is "13810.8 m".

First, we will find the deceleration of the motorcycle by using the first <em>equation of motion</em>:

v_f=v_it+at\\\\

where,

vi = initial velocity = (518 km/h)(\frac{1\ h}{3600\ s})(\frac{1000\ m}{1\ km}) = 143.89 m/s

vf = final veocity = 60 % of 143.89 m/s = (0.6)(143.89 m/s) = 86.33 m/s

a = deceleration = ?

t =time interval = 2 min = 120 s

Therefore,

86.33\ m/s = 143.89\ m/s + a(120\ s)\\\\a = \frac{86.33\ m/s - 143.89\ m/s}{120\ s}

a = -0.48 m/s²

Now, we will use the second <em>equation of motion </em>to find out the distance traveled (s):

s = v_it+\frac{1}{2}at^2\\\\s = (143.89\ m/s)(120\ s)+\frac{1}{2}(-0.48\ m/s^2)(120\ s)^2\\\\s = 17266.8\ m - 3456\ m

<u>s = 13810.8 m = 13.81 km</u>

<u />

Learn more about the equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion.

6 0
3 years ago
What is meant by heat energys​
ad-work [718]

Answer:

heat energy is the form of energy produced by heat

when we burn heat a type of enery is came

4 0
2 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
How many atoms are in the formula CH4? <br><br> A ) 2 <br><br> B ) 4 <br><br> C ) 5 <br><br> D ) 6
Rufina [12.5K]
B) 4 is the answer for atoms in CH4
6 0
2 years ago
Read 2 more answers
What happens if the breakdown voltage is exceeded.
s2008m [1.1K]

Answer:

this may help

Explanation:

maybe it will cause the eletric cause a fire that will mean it will spread till the fire men should come

7 0
2 years ago
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