Answer:
Torque = 35.60 N.m (rounded off to 3 significant figures.
Explanation:
Given details:
The mass of the rock on the left, ms = 2.25 kg
The total mass of the rocks, mp = 10.1 kg
The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m
(a) The torque produced by the pile of rock, T = F*rp = m*g*rp
Torque = 9.8*0.360*10.1 = 35.6328
Torque = 35.60 N.m (rounded off to 3 significant figures).
Answer:
1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%
Explanation:
For the first part, which is speed just before the crash, we can use energy conservation
Initial. Highest point
Em₀ = U = mg y
Final. Low point just before the crash
Emf = K = ½ m v²
Em₀ = Emf
m g y = ½ m v²
v = √ 2 g y
Let's calculate
v = √ (2 9.8 0.05)
v = 0.99 m / s
1) the moment before the crash is
p₀ = m v
p₀ = 0.221 0.99
p₀ = 0.219 kg m / s
After the collision, the car's speed is zero, so its moment is zero.
p = 0
2) change of momentum
Δp = p - p₀
Δp = 0- 0.219
Δp = -0.219 kg m / s
3) the reason is
Δp / p = 1
In percentage form it is 100%
Answer:
45000 K .
Explanation:
Given :
A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure
We need to find the temperature in which 1 litre of the same gas weigh 1 gram
in pressure 75 atm.
We know, by ideal gas equation :
Here , n is no of moles , 
Putting initial and final values and dividing them :
Hence , this is the required solution.
Answer:
A+B; 5√5 units, 341.57°
A-B; 5√5 units, 198.43°
B-A; 5√5 units, 18.43°
Explanation:
Given A = 5 units
By vector notation and the axis of A, it is represented as -5j
B = 3 × 5 = 15 units
Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B
(a) A + B = -5j + 15i
A+B = 15i -5j
|A+B| = √(15)²+(5)²
= 5√5 units
∆ = arctan(5/15) = 18.43°
The angle ∆ is generally used in the diagrams
∆= 18.43°
The direction of A+B is 341.57° based in the condition given (see attachment for diagrams
(b) A - B = -5j -15i
A-B = -15i -5j
|A-B|= √(15)²+(-5)²
|A-B| = √125
|A-B| = 5√5 units
The direction is 180+18.43°= 198.43°
See attachment for diagrams
(c) B-A = 15i -( -5j) = 15i + 5j
|B-A| = 5√5 units
The direction is 18.43°
See attachment for diagram
In order to change the direction and speed, a net external force is required. A net external force is an unbalanced force which will change the direction and gives the speed in the opposite direction. Hence, its an unbalanced force from the joey that pushes the car in the other direction due to which it the car starts to move back to Leah. Without, unbalanced force there is not change in the direction of the car's motion.
Hence, option B is correct.
