Here,
Height (S) = 2m
Gravity on mars (g) = 3.7m/s^2
Initial velocity (u) = 0 m/s^2
By the one of the formula of the motion,
S = ut + 1/2at^2
2 = 0 * t + 1/2*3.7*t^2
2 = 1.85t^2
t^2 = 2/1.85 = 1.081
t =1.03s
So, it will take 1.03s long..
Answer:
The magnitude of the electric field are
and ![1.09\times10^{4}\ N/C](https://tex.z-dn.net/?f=1.09%5Ctimes10%5E%7B4%7D%5C%20N%2FC)
Explanation:
Given that,
Radius of inner shell = 11.0 cm
Radius of outer shell = 14.0 cm
Charge on inner shell ![q_{inn}=3.50\times10^{-8}\ C](https://tex.z-dn.net/?f=q_%7Binn%7D%3D3.50%5Ctimes10%5E%7B-8%7D%5C%20C)
Charge on outer shell ![q_{out}=1.60\times10^{-8}\ C](https://tex.z-dn.net/?f=q_%7Bout%7D%3D1.60%5Ctimes10%5E%7B-8%7D%5C%20C)
Suppose, at r = 11.5 cm and at r = 20.5 cm
We need to calculate the magnitude of the electric field at r = 11.5 cm
Using formula of electric field
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Where, q = charge
k = constant
r = distance
Put the value into the formula
![E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes3.50%5Ctimes10%5E%7B-8%7D%7D%7B%2811.5%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E=2.38\times10^{4}\ N/C](https://tex.z-dn.net/?f=E%3D2.38%5Ctimes10%5E%7B4%7D%5C%20N%2FC)
The total charge enclosed by a radial distance 20.5 cm
The total charge is
![q=q_{inn}+q_{out}](https://tex.z-dn.net/?f=q%3Dq_%7Binn%7D%2Bq_%7Bout%7D)
Put the value into the formula
![q=3.50\times10^{-8}+1.60\times10^{-8}](https://tex.z-dn.net/?f=q%3D3.50%5Ctimes10%5E%7B-8%7D%2B1.60%5Ctimes10%5E%7B-8%7D)
![q=5.1\times10^{-8}\ C](https://tex.z-dn.net/?f=q%3D5.1%5Ctimes10%5E%7B-8%7D%5C%20C)
We need to calculate the magnitude of the electric field at r = 20.5 cm
Using formula of electric field
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Put the value into the formula
![E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes5.1%5Ctimes10%5E%7B-8%7D%7D%7B%2820.5%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E=1.09\times10^{4}\ N/C](https://tex.z-dn.net/?f=E%3D1.09%5Ctimes10%5E%7B4%7D%5C%20N%2FC)
Hence, The magnitude of the electric field are
and ![1.09\times10^{4}\ N/C](https://tex.z-dn.net/?f=1.09%5Ctimes10%5E%7B4%7D%5C%20N%2FC)
Ideal mechanical advantage doesn't take energy lost to friction into account. Explain why the mechanical advantage of a single fixed pulley is always 1. A single fixed pulley changes only the direction of the effort force. ... Energy transforms from the object supplying the force to the object being moved.
The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.
<h3>
Coefficient of the kinetic friction</h3>
The value of coefficient of kinetic friction is calculated as follows;
F - Ff = ma
F - μmgcosθ = ma
where;
- F is applied force
- μ is coefficient of kinetic friction
- m is mass of the wagon
- a is acceleration of the wagon
182 - μ(20 x 9.8 x cos30) = 20(2.5)
182 - 169.74μ = 50
182 - 50 = 169.74μ
132 = 169.74μ
μ = 132/169.74
μ = 0.78
Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.
Learn more about coefficient of friction here: brainly.com/question/20241845
Answer:
The Statement is wrong because the reverse is the case as it is the kinetic energy that is being transformed to gravitational potential energy.
Explanation:
As your friend throws the baseball into the air the ball gains an initial velocity (u) and this makes the Kinetic energy to be equal to
![KE = \frac{1}{2} mu^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mu%5E2)
Here m is the mass of the baseball
Now as this ball moves further upward the that velocity it gained reduce due to the gravitational force and this in turn reduces the kinetic energy of the ball and this kinetic energy lost is being converted to gravitational potential energy which is mathematically represented as (m×g×h)
as energy can not be destroyed but converted to a different form according to the first law of thermodynamics
Looking a the formula for gravitational potential energy we see that the higher the ball goes the grater the gravitational potential energy.