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Bingel [31]
3 years ago
7

A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizonta

l distance between the two towers is 20 meters. With what horizontal velocity should the ball be imparted so that it lands on the rooftop of the second building?
Physics
1 answer:
I am Lyosha [343]3 years ago
7 0
-- During the time the ball is flying from the high roof to the low roof,
it's going to fall (100-25) = 75 meters.
How long does it take an object dropped from rest to fall 75 meters ?

Distance = (1/2) · (gravity) · (time)²

75 m = (4.9 m/s²) · (time)²

Time² = (75 m) / (4.9 m/s²)
Time² = 15.31 sec²
Time = √(15.31 sec²)  =  3.91 seconds 

So the ball has to cover the horizontal distance of 20 meters 
in 3.91 seconds.

Distance = (speed) · (time)

20 m = (speed) · (3.91 sec)

Speed = (20 m) / (3.91 sec)

Speed  =  5.11 m/s
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goldfiish [28.3K]

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3 years ago
A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-
Alisiya [41]
First we need to convert the angular speed from rpm to rad/s. Keeping in mind that 
1 rev= 2 \pi rad
1 min = 60 s
the angular speed is
\omega = 7.18  \frac{rev}{min} \cdot  \frac{2 \pi}{60} = 0.75 rad/s

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s
3 0
4 years ago
Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma
Sloan [31]

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance (C_{HP}) is given by

C_{HP} = \dfrac{Q_{H}}{W_{in}}

where, Q_{H} is the magnitude of heat transfer between cyclic device and    high-temperature medium at temperature T_{H} and W_{in} is the required input and is given by W_{in} = Q_{H} - Q_{L}, Q_{L} being magnitude of heat transfer between cyclic device and low-temperature T_{L}. Therefore, from above equation we can write,

&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}

Given, T_{L} = 460 K and T_{H} = 540 K. So,  the minimum work per unit heat transfer is given by

\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15

8 0
4 years ago
While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000
Minchanka [31]

Answer:

a

    The hiker (you ) is  200 m below his/her(your) starting point

b

   The resultant displacement in the north east direction is

a  = 6562.0 \  m

    The resultant displacement in vertical direction (i.e the altitude change )

  b =6503.1 \  m

Explanation:

From the question we are told that  

  The displacement in the morning is  S_{morning} =  (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)

   The displacement in the afternoon is  S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)

Generally the direction west is negative , the direction east is positive

                 the direction south is negative , the direction north is  positive

resultant displacement  is mathematically evaluated as  

    (2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east

     (4000\ m\ north)  + (2500 \ m ,\ north) = 6500  \ m ,\ north

     (100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m

From the above calculation we see that at the end of the hiking the hiker (you) is  200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

      a  =  \sqrt{900^2 + 6500^2}

=>     a  = 6562.0 \  m

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

      b = \sqrt{6500^2 +(-200)^2  }

=>   b =6503.1 \  m

   

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horrorfan [7]
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6 0
4 years ago
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