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Len [333]
3 years ago
6

a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of

the rocket and hits the ground 6.00 seconds later. what was the rockets acceleration?
Physics
1 answer:
Papessa [141]3 years ago
6 0

The initial speed of the bolt is not 58.86 m/s.  

Let a be the acceleration of the rocket.  

During the 4 sec lift off, the rocket has reached a height of  

h = (1/2)*a*t^2  

with t=4,  

h = (1/2)*a^16  

h = 8*a  

Its velocity at 4 sec is  

v = t*a  

v = 4*a  

The initial velocity of the bolt is thus 4*a.  

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,  

h = (1/2)*g*t^2 + V0*t  

Substituting h0=8*a, t=6 and V0=-4*a into it,  

8*a = (1/2)*g*36 - 4*a*6  

Solving for a  

a = 5.52 m/s^2

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If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
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Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

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Solving the two expressions simultaneously for x we will have;

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Cross multiply to get x

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6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

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A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

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q is the distance of the image from the lens

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p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

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q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

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