Answer:
Explanation:
a ) Height to be cleared = 5 - 1.6 = 3.4 m
Horizontal distance to be cleared = 5 m .
angle of throw = 56°
here y = 3.4 , x = 5 , θ = 56
equation of trajectory
y = x tanθ - 1/2 g ( x/ucosθ)²
3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²
3.4 = 7.4 - 122.5 / .3125u²
122.5 / .3125u² = 4
u² = 98
u = 9.9 m /s
Range = u² sin 2 x 56 / g
= 9.9 x 9.9 x .927 / 9.8
= 9.27 m
horizontal distance be-yond the fence will the rock land on the ground
= 9.27 - 5
= 4.27 m
Search it up and there’s a video of how to make spectroscope.
Answer: Total work done on the block is 3670.5 Joules.
Step by step:
Work done:
![W = F.d.\cos\theta](https://tex.z-dn.net/?f=W%20%3D%20F.d.%5Ccos%5Ctheta)
With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)
![W = F.d](https://tex.z-dn.net/?f=W%20%3D%20F.d)
Given:
F = 75 N
m = 31.8 kg
Final velocity ![v_f = 15.3 \frac{m}{s}](https://tex.z-dn.net/?f=v_f%20%3D%2015.3%20%5Cfrac%7Bm%7D%7Bs%7D)
In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:
![F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\](https://tex.z-dn.net/?f=F%20%3D%20ma%5C%5C%5Cimplies%20a%3D%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7B75%20N%20%7D%7B31.8%20kg%7D%5Capprox%202.36%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%5C%5C)
Now we can determine the displacement from the following formula:
![d = \frac{1}{2}a^2+v_0t+d_0](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B1%7D%7B2%7Da%5E2%2Bv_0t%2Bd_0)
Here, the initial displacement is 0 and initial velocity is also 0 (at rest):
![d = \frac{1}{2}at^2\\](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C)
Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:
![v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s](https://tex.z-dn.net/?f=v_f%20%3D%20a%5Ccdot%20t_f%5C%5C%5Cimplies%20t_f%20%3D%20%5Cfrac%7Bv_f%7D%7Ba%7D%20%3D%20%5Cfrac%7B15.2%20%5Cfrac%7Bm%7D%7Bs%7D%7D%7B2.36%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%7D%5Capprox%206.44%20s)
(using vf as final velocity, and tf as final time)
So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:
![d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%3D%5Cfrac%7B1%7D%7B2%7D2.36%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%5Ccdot%206.44%5E2%20s%5E2%20%5Capprox%2048.94%20m)
and the corresponding work:
![W = F\cdot d=75 N\cdot 48.94 m =3670.5J](https://tex.z-dn.net/?f=W%20%3D%20F%5Ccdot%20d%3D75%20N%5Ccdot%2048.94%20m%20%3D3670.5J)
Answer:
The maximum electric field ![E_{max}= 0.132V/m](https://tex.z-dn.net/?f=E_%7Bmax%7D%3D%200.132V%2Fm)
Explanation:
From the question we are told that
The diameter is ![d = 3.3 cm = \frac{3.3}{100} = 0.033m](https://tex.z-dn.net/?f=d%20%3D%203.3%20cm%20%3D%20%5Cfrac%7B3.3%7D%7B100%7D%20%3D%200.033m)
The magnetic field of the cylinder is
The frequency is ![f = 15Hz](https://tex.z-dn.net/?f=f%20%3D%2015Hz)
The radial distance is ![d_r = 1.4cm = \frac{1.4}{100} = 0.014m](https://tex.z-dn.net/?f=d_r%20%3D%201.4cm%20%3D%20%5Cfrac%7B1.4%7D%7B100%7D%20%20%3D%200.014m)
This magnetic field can be represented mathematically as
![B(t) = B_i + B_1sin (wt + \o_i)](https://tex.z-dn.net/?f=B%28t%29%20%3D%20B_i%20%2B%20B_1sin%20%28wt%20%2B%20%5Co_i%29)
The initial magnetic field is the average between the variation of the magnetic field which is represented as
![B_i = \frac{30 + 29.6}{2}](https://tex.z-dn.net/?f=B_i%20%3D%20%5Cfrac%7B30%20%2B%2029.6%7D%7B2%7D)
![= 29.8T](https://tex.z-dn.net/?f=%3D%20%2029.8T)
Then
is the amplitude of the resultant field is mathematically evaluated as
![B_1 = \frac{30.0 - 29.6}{2}](https://tex.z-dn.net/?f=B_1%20%3D%20%5Cfrac%7B30.0%20-%2029.6%7D%7B2%7D)
![= 0.200T](https://tex.z-dn.net/?f=%3D%200.200T)
The electric field induced can be represented mathematically as
![E = \frac{1}{2} [\frac{dB }{dt} ]d_r](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20%5B%5Cfrac%7BdB%20%7D%7Bdt%7D%20%20%5Dd_r)
![= \frac{d_r}{2} \frac{d}{dt} (B_i + B_1 sin (wt + \o_o))](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bd_r%7D%7B2%7D%20%5Cfrac%7Bd%7D%7Bdt%7D%20%28B_i%20%2B%20B_1%20sin%20%28wt%20%2B%20%5Co_o%29%29)
![= \frac{1}{2} (B wr cos (wt + \o_o))](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28B%20wr%20cos%20%28wt%20%2B%20%5Co_o%29%29)
At maximum electric field ![cos (wt + \o_o) = 1](https://tex.z-dn.net/?f=cos%20%28wt%20%2B%20%5Co_o%29%20%20%3D%201)
![E_{max} = \frac{1}{2} B_1 wd_r](https://tex.z-dn.net/?f=E_%7Bmax%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20B_1%20wd_r)
![E_{max} = \frac{1}{2} B_1 2 \pi f d_r](https://tex.z-dn.net/?f=E_%7Bmax%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20B_1%202%20%5Cpi%20f%20d_r)
![= \frac{1}{2} (0.200) (2 \pi (15 ))(0.014)](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B2%7D%20%280.200%29%20%282%20%5Cpi%20%2815%20%29%29%280.014%29)
![E_{max}= 0.132V/m](https://tex.z-dn.net/?f=E_%7Bmax%7D%3D%200.132V%2Fm)
Answer:
we approach a maximum or minimum the values of the ordinate are closer and closer and when passing this point the values change their trend
Explanation:
The reason for this process occurs because as we approach a maximum or minimum the values of the ordinate are closer and closer and when passing this point the values change their trend if they were rising, they begin to fall and if they were falling they begin to rise. Therefore the maximum point is a point of inflection of the curve since its trend changes.
Another way of looking at this process is that mathematically the point where there is a maximum or a minimum corresponds to the point where the first derivative is equal to zero, this is the slope of the line is horizontal, so the points before after correspond to values with slope of different sign.