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Mila [183]
2 years ago
11

In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9

.0 m above him, releases a ball from rest. While still in flight, the acrobat catches the ball just as it reaches him.
If he left the trampoline with a speed of 5.6 m/s, how long is he in the air before he catches the ball? (express your answer in second)
Physics
1 answer:
mariarad [96]2 years ago
6 0

Answer:

1.6 s

Explanation:

Let after time t both ball and acrobat meet in the mid air .

Distance traveled by acrobat

h₁ = u t -.5 g t²

= 5.6 t - 4.9 t²

Distance traveled by ball in time t , when it is released with initial velocity zero

h₂ = .5 g t²

= 4.9 t²

Now h₁ + h₂ = 9 m

5.6t - 4.9 t² + 4.9 t² = 9

5.6 t = 9

t =  \frac{9}{5.6}

1.6 s

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2 years ago
An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48
kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

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Er = remaining energy [J]

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3 0
2 years ago
At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w
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Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s 
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15,000/5,000 = 3 
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6 0
2 years ago
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a
Svetllana [295]

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 <em>u </em>s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 <em>u </em>s  = 75 <em>n</em> s

TL = T - Th = 100 <em>n</em>s - 75 <em>n</em> s = 25 <em>n</em> s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

8 0
2 years ago
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