Complete question:
A hydraulic jack is used to lift a car by applying a force of 120N at the pump piston, if the area of the ram and pump piston are 100cm squared and 1m squared respectively. What is the weight of the car?
Answer:
the weight of the car is 1.2 N
Explanation:
Given;
applied force, F₁ = 120 N
area of the effort (pump), A₁ = 1 m²
area of the load (ram), A₂ = 100 cm² = 1 x 10⁻² m²
let the weight of the car = F₂
The applied pressure is constant and the following equations can be used to calculate the weight of the car.
![P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\F_2 = \frac{F_1 \times A_2}{A_1} \\\\F_2 = \frac{120 \times (1\times 10^{-2})}{1} \\\\F_2 = 1.2 \ N](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BF_1%7D%7BA_1%7D%20%3D%20%5Cfrac%7BF_2%7D%7BA_2%7D%20%5C%5C%5C%5CF_2%20%3D%20%5Cfrac%7BF_1%20%5Ctimes%20A_2%7D%7BA_1%7D%20%5C%5C%5C%5CF_2%20%3D%20%5Cfrac%7B120%20%5Ctimes%20%281%5Ctimes%2010%5E%7B-2%7D%29%7D%7B1%7D%20%5C%5C%5C%5CF_2%20%3D%201.2%20%5C%20N)
Therefore, the weight of the car is 1.2 N
Answer:
a) 627.84 Joules
b) 117.72 Joules
c) 1255.68 Joules
Explanation:
<em>(See figure 1)</em>
The gravitational potential energy relative to the child’s lowest position is:
(1)
with h the vertical distance of the swing from the lowest position, m the mass of the child and g the acceleration of gravity.
a) When the ropes are horizontal, the swing is at 1.60 m from the lowest position, so by (1):
![U_{g}=(40)(9.81)(1.60)=627.84\,J](https://tex.z-dn.net/?f=%20U_%7Bg%7D%3D%2840%29%289.81%29%281.60%29%3D627.84%5C%2CJ)
b) When the ropes make a 36.0◦ angle with the vertical, the swing is at a distance d from the lowest position, we should use trigonometric relations to find that distance. By figure 2 we have a right triangle with adjacent side
and hypotenuse 1.60, so using the trigonometric relation
we can solve for d:
![d=1.60-1.60*\cos(36)\approx0.30\,m](https://tex.z-dn.net/?f=%20d%3D1.60-1.60%2A%5Ccos%2836%29%5Capprox0.30%5C%2Cm%20)
Using d on (1):
![U_{g}=(40)(9.81)(0.30)=117.72\,J](https://tex.z-dn.net/?f=%20U_%7Bg%7D%3D%2840%29%289.81%29%280.30%29%3D117.72%5C%2CJ%20)
c) Note that at the bottom of the circular arc the distance of the swing relative to the lowest position is two times the length of the rope, so h=3.20 m, using this on (1):
![U_{g}=(40)(9.81)(3.20)=1255.68\,J](https://tex.z-dn.net/?f=U_%7Bg%7D%3D%2840%29%289.81%29%283.20%29%3D1255.68%5C%2CJ%20)
To answer this question, we need to find the force acting on the two objects.
f=ma
f=(6)(2)
f=12 newtons
So, now we can determine the acceleration of the second object using the same formula!
f=ma
12=(4)a
3=a
So the acceleration of the smaller object is 3 meters per second squared!
Answer:
Distance travelled, d = 0.21 m
Explanation:
It is given that,
Initial velocity of electron, u = 500,000 m/s
Acceleration of the electron, a = 500,000,000,000 m/s²
Final velocity of the electron, v = 675,000 m/s
We need to find the distance travelled by the electron. Let distance travelled is s. Using third equation of motion as :
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![s=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![s=\dfrac{(675000\ m/s)^2-(500000\ m/s^2)^2}{2\times 500000000000\ m\s^2}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B%28675000%5C%20m%2Fs%29%5E2-%28500000%5C%20m%2Fs%5E2%29%5E2%7D%7B2%5Ctimes%20500000000000%5C%20m%5Cs%5E2%7D)
s = 0.205 m
or
s = 0.21 m
So, the electron will travel a distance of 0.21 meters. Hence, this is the required solution.