Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = 
m
Width of the window = 
m
Area of the window = 
Difference in air pressure = Inside pressure - Outside pressure
= 
atm = 
atm
Conversion of the pressure in its SI unit.
⇒ 
atm = 
Pa
⇒ atm = 
Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒ 
⇒ 
⇒ Plugging the values.
⇒ 
⇒ 
Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
Answer:
0.5kg
Explanation:
Given parameters:
Potential energy = 147J
Height = 30m
Unknown:
Mass of the bird = ?
Solution:
Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;
P.E = mgH
m is the mass
g is the acceleration due to gravity = 9.8m/s²
H is the height
147 = m x 9.8 x 30
m = 0.5kg
Answer:
18.63 N
Explanation:
Assuming that the sum of torques are equal
Στ = Iα
First wheel
Στ = 5 * 0.51 = 3 * (0.51)² * α
On making α subject of formula, we have
α = 2.55 / 0.7803
α = 3.27
If we make the α of each one equal to each other so that
5 / (3 * 0.51) = F2 / (3 * 1.9)
solve for F2 by making F2 the subject of the formula, we have
F2 = (3 * 1.9 * 5) / (3 * 0.51)
F2 = 28.5 / 1.53
F2 = 18.63 N
Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.
Answer:
A. Two tennis balls that are near each other
Explanation:
The formula for gravitational force (F) between two objects is
where m₁ and m₂ are the masses of the two objects, d is the distance between their centres, and G is the gravitational constant.
Thus, two objects that are far from each other will have a smaller gravitational force. We can eliminate Options C and D.
If the objects are at the same distance, those with the smaller mass will have a smaller force.
The mass of a tennis ball is 57 g.
The mass of a soccer ball is 430 g.
Two tennis balls that are near each other will have a smaller gravitational attraction.
Answer:
The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
Explanation:
Electric potential is given as;
V = E*r
where;
E is the electric field strength, = kq/r²
V = ( kq/r²)*r
V = kq/r
k is coulomb's constant = 8.99 X 10⁹ Nm²/C²
q is the charge of the particles = 1.6 X 10⁻¹⁹ C
r is the distance between the particles = 859 nm
At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm
V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)
V = 3.349 X 10⁻³ Volts
Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
