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creativ13 [48]
3 years ago
6

66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st

raight up with an initial velocity of 15.0 m/s.
Physics
1 answer:
kozerog [31]3 years ago
3 0

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

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Balance the following chemical equation:<br> CO2 + VH20 - CH1206+ voz<br> [<br> ~<br> CO₂+
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Answer:

Explanation:

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3 0
3 years ago
Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a
wel

Answer:

a) v=7.32m/s

b) \alpha =-35º

c) ΔK=-1094.62J

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

p_{1x}=p_{2x}

To simplify the problem, lets name D for Daniel and R for Rebecca

a) p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}

Since Daniel's initial velocity is 0

m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}

v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}

v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s

Now, lets analyze the movement in the vertical direction

p_{1y}=p_{2y}

Since p_{1y}=0

0=m_{D}v_{D2y}+m_{R}v_{R2y}

v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s

Now, we can find the magnitude of Daniel's velocity after de collision

v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35º

c) The change in the total kinetic energy is:

ΔK=K_{2}-K_{1}

ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
2 years ago
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