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LUCKY_DIMON [66]
3 years ago
7

Samantha is 1.1 m tall on her eleventh birthday and 1.21 m tall on her twelfth birthday. By what factor has her height increased

?
Physics
1 answer:
Maurinko [17]3 years ago
7 0
Well to figure this out you can simply get how tall she is now which is 1.21 m tall and subtract how tall she used to be which is 1.1 from that. 
If you subtract them you should have got 0.11 m making the total that she grew to be 0.11. 
Hope I Helped.:)
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When does the potential energy of a swinging pendulum increase the most?
xxTIMURxx [149]
Maybe this can help:

https://blogs.bu.edu/ggarber/interlace/pendulum/energy-in-a-pendulum/
3 0
3 years ago
An object is moving at a velocity of 23 m/s. It accelerates to a velocity of 85 m/s over a time of 8.3 s. What acceleration did
Bess [88]
\frac{v \: final \: - v \: initial }{time} = \: a
\\ \frac{85 - 23}{8.3} = 7.467 \: \frac{m}{ {s}^{2} }
7 0
3 years ago
Read 2 more answers
A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0
ivolga24 [154]
Since you solved a and b I will start with part c.
Part C
To answer this question we need to find zeros of a velocity function:
v(t)=0.04t^3-0.06t^2
We can factor this polynomial:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)
Now it's pretty easy to find zeros. This function will be equal to zero when any of the factors are equal zero. 
t^2=0;\\ 0.04t-0.06=0
We solve these two equations and we get our zeros:
t_1=0; t_2=\frac{3}{2}
The particle is at rest at t=0 and t=3/2.
Part D
To solve this we need to determine when our velocity function is greater than zero. We will use factored form. 
We determine when each factor is greater than zero and with that information, we build the following table:
\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}
We can see, from the table, that our function is positive when - \infty < t and t>3/2.
That is the range in which particle is moving in positive direction.
Part E
We know that distance traveled is given with:
s(t)=0.01t^4 - 0.02t^3
We simply plug in t=12 to find total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft
Part F
We know that acceleration is defined as a rate of change of velocity.
We find acceleration by taking the first derivative of velocity with respect to time.
a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t
To find acceleration after 1 second we simply plug in t=1s in above equation:
a(1)=0.12-0.12=0


7 0
3 years ago
Two sources of light of wavelength 720 nm are 10 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for
julia-pushkina [17]

Answer:

The value is  y  =  0.00732 \ m    

Explanation:

From the question we are told that

    The wavelength of each source is  \lambda  = 720 \ nm =  720 *10^{-9} \  m

     The distance from the pinhole D = 10 \ m

     The diameter is d =  1.2 mm = \frac{1.2}{1000} = 0.0012 \ m

Generally from Rayleigh's criterion  we have that the distance between the sources of light for their diffraction patterns is mathematically represented as '

        y  =  \frac{ 1.22 *  \lambda *  D}{d}

=>      y  =  \frac{ 1.22 * 720 *10^{-9}*   10 }{ 0.0012}

=>      y  =  0.00732 \ m      

6 0
3 years ago
A 2.6 kg rock is dropped from a height of 10 m. With what speed will it strike the ground. Ignore air resistance. Solve using co
timurjin [86]

Answer:

mgh =  \frac{1}{2} m {v}^{2}  \\ v =  \sqrt{2gh}  \\  =  \sqrt{2 \times 9.8 \times 10}  \: m. {s}^{ - 1}

5 0
3 years ago
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