Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J
ANSWER
![\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%2A%7D%204.67%2A10%5E%7B-9%7D%5Ctext%7B%20%7DN%20%5Cend%7Bequation%2A%7D)
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
![F=\frac{GMm}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D)
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
![\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20F%3D%5Cfrac%7B6.67430%2A10%5E%7B-11%7D%2A70%2A1%7D%7B1%5E2%7D%20%5C%5C%20%20%5C%5C%20F%3D4.67%2A10%5E%7B-9%7D%5Ctext%7B%20%7DN%20%5Cend%7Bgathered%7D)
That is the answer.
Answer:
m = 81281.5 pounds.
Explanation:
Given that,
Force, F = 73 kN
Acceleration of the space shuttle, a = 16000 mi/h²
1 miles/h² = 0.0001241 m/s2
16000 mi/h² = 1.98 m/s²
We need to find the mass of the spacecraft.
According to Newton's second law,
F = ma
m is mass of the spacecraft
![m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BF%7D%7Ba%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B73%5Ctimes%2010%5E3%5C%20N%7D%7B1.98%5C%20m%2Fs%5E2%7D%5C%5C%5C%5Cm%3D36868.68%5C%20kg)
Since, 1 kg = 2.20462 pounds
m = 81281.5 pounds
Hence, the mass of the spacecraft is 81281.5 pounds.
PART a)
As we know that gravitational potential energy is given by the formula
![U = -\frac{Gm_1m_2}{r}](https://tex.z-dn.net/?f=U%20%3D%20-%5Cfrac%7BGm_1m_2%7D%7Br%7D)
here we can see that gravitational potential energy inversely varies with the distance
so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position
So gravitational potential energy is minimum at the nearest point and maximum at the farthest point
PART b)
Since we know that sum of kinetic energy and potential energy is constant here
so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum
So here speed is maximum at the nearest point
Part C)
since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance
Answer:
(a) 45 micro coulomb
(b) 6 micro Coulomb
Explanation:
C = 3 micro Farad = 3 x 10^-6 Farad
V = 15 V
(a) q = C x V
where, q be the charge.
q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb
(b)
V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad
q = C x V
where, q be the charge.
q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb