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blondinia [14]
3 years ago
12

Can some answer 7 a and b please

Physics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

t = 15 \ sec

Plugging the values we have.

V_{f} =V_{i} + at

V_{f} =0.8 + 0.06 \times 15

V_{f} =0.8+ 0.9

V_{f} =1.7 m/s

Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

s= 18.75\ m

So Mr.Comer will travel a distance of s= 18.75\ m.

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Good afternoon.


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A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,
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For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem: 

s=1.80m when t=0.245s 

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation): 

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

v=8.549m/s 

<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t) 

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like 

0=(v)+(a)(t) 

Substituting and solving for t: 

0=(8.549m/s)+(-9.81m/s^2)(t) 

t=0.8714s 

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m</span>

 

Answer:

<span>1.925m</span>

3 0
3 years ago
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