Answer:
The time period of the empty car will be "1.00 s".
Explanation:
The given values in the question will be:
Mass,
m = 250 kg
Loaded car's time period will be:
T = 1.08 s
Shock absorbers compression,
x = 4 cm
or,
= 0.04 m
Now,
Weight of passengers will be:
⇒ ![F=mg](https://tex.z-dn.net/?f=F%3Dmg)
![=250\times 9.8](https://tex.z-dn.net/?f=%3D250%5Ctimes%209.8)
![=2450 \ N](https://tex.z-dn.net/?f=%3D2450%20%5C%20N)
The spring constant of shock absorbers will be:
⇒ ![k=\frac{F}{x}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BF%7D%7Bx%7D)
![=\frac{2450}{0.04}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2450%7D%7B0.04%7D)
![=61.250 \ N/m](https://tex.z-dn.net/?f=%3D61.250%20%5C%20N%2Fm)
As we know,
Time period, ![T = 2 \pi\sqrt{\frac{M}{k} }](https://tex.z-dn.net/?f=T%20%3D%202%20%5Cpi%5Csqrt%7B%5Cfrac%7BM%7D%7Bk%7D%20%7D)
On substituting the values, we get
![1.08=2\pi \sqrt{\frac{M}{61250} }](https://tex.z-dn.net/?f=1.08%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7BM%7D%7B61250%7D%20%7D)
![\frac{M}{61250}=0.02955](https://tex.z-dn.net/?f=%5Cfrac%7BM%7D%7B61250%7D%3D0.02955)
![M=0.02955\times 61250](https://tex.z-dn.net/?f=M%3D0.02955%5Ctimes%2061250)
(Total mass of car as well as its passengers)
Now,
The mass of the empty car will be:
⇒ ![m'=M-m](https://tex.z-dn.net/?f=m%27%3DM-m)
![=1809.6-250](https://tex.z-dn.net/?f=%3D1809.6-250)
![=1559.6 \ kg](https://tex.z-dn.net/?f=%3D1559.6%20%5C%20kg)
hence,
The time period of empty car will be:
⇒ ![T'=2\pi\sqrt{\frac{m'}{k} }](https://tex.z-dn.net/?f=T%27%3D2%5Cpi%5Csqrt%7B%5Cfrac%7Bm%27%7D%7Bk%7D%20%7D)
![=2\pi\sqrt{\frac{1559.6}{61250} }](https://tex.z-dn.net/?f=%3D2%5Cpi%5Csqrt%7B%5Cfrac%7B1559.6%7D%7B61250%7D%20%7D)
![=2\pi \sqrt{0.0254}](https://tex.z-dn.net/?f=%3D2%5Cpi%20%5Csqrt%7B0.0254%7D)
![=1.003 \ s](https://tex.z-dn.net/?f=%3D1.003%20%5C%20s)
or,
![=1.00 \ s](https://tex.z-dn.net/?f=%3D1.00%20%5C%20s)
Answer:
See below
Explanation:
With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings
V = IR
I = V/R = 5 / (10+5+5) = .25 A
b) With S1 closed 5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm
then the series circuit current becomes
5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps
ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3 .273 * 10 / 15 =.182 amps
Answer:
Part a)
![\lambda = 300 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20300%20m)
Part b)
![E = 2.7 N/C](https://tex.z-dn.net/?f=E%20%3D%202.7%20N%2FC)
Part c)
![I = 9.68 \times 10^{-3} W/m^2](https://tex.z-dn.net/?f=I%20%3D%209.68%20%5Ctimes%2010%5E%7B-3%7D%20W%2Fm%5E2)
![P = 3.22 \times 10^{-11} N/m^2](https://tex.z-dn.net/?f=P%20%3D%203.22%20%5Ctimes%2010%5E%7B-11%7D%20N%2Fm%5E2)
Explanation:
Part a)
As we know that frequency = 1 MHz
speed of electromagnetic wave is same as speed of light
So the wavelength is given as
![\lambda = \frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bc%7D%7Bf%7D)
![\lambda = \frac{3\times 10^8}{1\times 10^6}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B3%5Ctimes%2010%5E8%7D%7B1%5Ctimes%2010%5E6%7D)
![\lambda = 300 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20300%20m)
Part b)
As we know the relation between electric field and magnetic field
![E = Bc](https://tex.z-dn.net/?f=E%20%3D%20Bc)
![E = (9 \times 10^{-9})(3\times 10^8)](https://tex.z-dn.net/?f=E%20%3D%20%289%20%5Ctimes%2010%5E%7B-9%7D%29%283%5Ctimes%2010%5E8%29)
![E = 2.7 N/C](https://tex.z-dn.net/?f=E%20%3D%202.7%20N%2FC)
Part c)
Intensity of wave is given as
![I = \frac{1}{2}\epsilon_0E^2c](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cepsilon_0E%5E2c)
![I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%288.85%20%5Ctimes%2010%5E%7B-12%7D%29%282.7%29%5E2%283%5Ctimes%2010%5E8%29)
![I = 9.68 \times 10^{-3} W/m^2](https://tex.z-dn.net/?f=I%20%3D%209.68%20%5Ctimes%2010%5E%7B-3%7D%20W%2Fm%5E2)
Pressure is defined as ratio of intensity and speed
![P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BI%7D%7Bc%7D%20%3D%20%5Cfrac%7B9.68%5Ctimes%2010%5E%7B-3%7D%7D%7B3%5Ctimes%2010%5E8%7D)
![P = 3.22 \times 10^{-11} N/m^2](https://tex.z-dn.net/?f=P%20%3D%203.22%20%5Ctimes%2010%5E%7B-11%7D%20N%2Fm%5E2)
First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton
Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons
Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction)
Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg
Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2
Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters
Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec
Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec
"The marble moved 30 cm north in 6 seconds" is the one example among the following choices given in the question that <span>provides a complete scientific description of an object in motion. The correct option among all the options that are given in the question is the third option or option "C". I hope the answer has helped you.</span>