Answer:
![v = 12.57 m/s](https://tex.z-dn.net/?f=v%20%3D%2012.57%20m%2Fs)
Explanation:
As we know that the radius of the circular motion is given as
![R = 50.00 cm](https://tex.z-dn.net/?f=R%20%3D%2050.00%20cm)
time period of the motion is given as
![T = 0.2500 s](https://tex.z-dn.net/?f=T%20%3D%200.2500%20s)
now we know that it is moving with uniform speed
so it is given as
![v = \frac{2\pi R}{T}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%20R%7D%7BT%7D)
now plug in all data
![v = \frac{2\pi(0.50)}{0.25}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%280.50%29%7D%7B0.25%7D)
![v = 12.57 m/s](https://tex.z-dn.net/?f=v%20%3D%2012.57%20m%2Fs)
Answer:
a ![W_{3.5} = 490 \ J](https://tex.z-dn.net/?f=W_%7B3.5%7D%20%3D%20490%20%5C%20%20J)
b
Explanation:
Generally the force constant is mathematically represented as
![k = \frac{F}{x}](https://tex.z-dn.net/?f=k%20%20%3D%20%5Cfrac%7BF%7D%7Bx%7D)
substituting values given in the question
=> ![k = \frac{16}{0.2}](https://tex.z-dn.net/?f=k%20%20%3D%20%5Cfrac%7B16%7D%7B0.2%7D)
=> ![k = 80 \ N /m](https://tex.z-dn.net/?f=k%20%20%3D%20%2080%20%5C%20N%20%2Fm)
Generally the workdone in stretching the spring 3.5 m is mathematically represented as
![W_{3.5} = \frac{1}{2} * k * (3.5)^2](https://tex.z-dn.net/?f=W_%7B3.5%7D%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20%283.5%29%5E2)
=> ![W_{3.5} = \frac{1}{2} * 80 * (3.5)^2](https://tex.z-dn.net/?f=W_%7B3.5%7D%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%2080%20%20%2A%20%20%283.5%29%5E2)
=> ![W_{3.5} = 490 \ J](https://tex.z-dn.net/?f=W_%7B3.5%7D%20%3D%20490%20%5C%20%20J)
Generally the workdone in compressing the spring 2.5 m is mathematically represented as
=>
=>
Answer:
The electric flux is ![280\ \rm N.m^2/C](https://tex.z-dn.net/?f=280%5C%20%5Crm%20N.m%5E2%2FC)
Explanation:
Given:
- Radius of the disc R=0.50 m
- Angle made by disk with the horizontal
![\theta=30^\circ](https://tex.z-dn.net/?f=%5Ctheta%3D30%5E%5Ccirc)
- Magnitude of the electric Field
![E=713.0\ \rm N/C](https://tex.z-dn.net/?f=E%3D713.0%5C%20%5Crm%20N%2FC)
The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows
![\phi=\int E.dA](https://tex.z-dn.net/?f=%5Cphi%3D%5Cint%20E.dA%20)
where
is the total Electric Flux- E is the Electric Field
- dA is the Area through which the electric flux is to be calculated.
Now according to question we have
![=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C](https://tex.z-dn.net/?f=%3DEA%5Ccos%5Ctheta%20%5C%5C%3D713%5Ctimes%203.14%5Ctimes%200.5%5E2%20%5Ctimes%20%5Ccos60%5E%5Ccirc%5C%5C%3D280%5C%20%5Crm%20N.m%5E2%2FC)
Hence the electric flux is calculated.
Answer:
See the explanation below
Explanation:
The pressure is defined as the product of the density of the liquid by the gravitational acceleration by the height, and can be easily calculated by means of the following equation.
![P=Ro*g*h](https://tex.z-dn.net/?f=P%3DRo%2Ag%2Ah)
where:
Ro = density of the fluid [kg/m³]
g = gravity acceleration = 9.81 [m/s²]
h = elevation [m]
In this way we can understand that the greater pressure is achieved by means of the height of the liquid, that is, as long as the fluid has more height, greater pressure will be achieved at the bottom.
Therefore in order of decreasing will be
The largest pressure with the largest height of the liquid, container B. The next is obtained with container D, the next with container A and the lowest pressure with container C.
The pressure decreases as we go from the container B - D - A - C
The rotation of Earth is equivalent to one day which is comprised of 24 hours. To determine the number of miles in Earth's circumference, one simply have to multiply the given rate by the appropriate conversion factor and dimensional analysis. This is shown below.
C = (1038 mi/h)(24 h/1 day)
C = 24,912 miles
From the given choices, the nearest value would have to be 20,000 mile. The answer is the second choice.