Two explanations might include two sets of colorless solutions that might yield a gas. What comes to mind is the production of carbon dioxide which could come from a soluble carbonate, such as sodium carbonate solution, and an acid, such as vinegar or hydrochloric acid. A second possibility is a gas such as ammonia. This could be produced by the reaction of a soluble ammonium compound, such as ammonium chloride, and a base such as sodium hydroxide.
Answer:
The mass of the block, M =T/(3a +g) Kg
Explanation:
Given,
The upward acceleration of the block a = 3a
The constant force acting on the block, F₀ = Ma = 3Ma
The mass of the block, M = ?
In an Atwood's machine, the upward force of the block is given by the relation
Ma = T - Mg
M x 3a = T - Ma
3Ma + Mg = T
M = T/(3a +g) Kg
Where 'T' is the tension of the string.
Hence, the mass of the block in Atwood's machine is, M = T/(3a +g) Kg
Answer:
a) The distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object is 5 m/s².
Explanation:
a) The distance can be found using the equation of gravitational force:
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
M: is the Earth's mass = 5.97x10²⁴ kg
m: is the object's mass = 0.4 kg
F: is the force or the weight = 2.0 N
r: is the distance =?
The distance is:
Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object can be calculated knowing the weight:
Where:
W: is the weight = 2 N
a: is the initial acceleration =?
Therefore, the initial acceleration of the object is 5 m/s².
I hope it helps you!
Answer:
The impulse received by the ball is - 1.561 kg.m/s
Explanation:
Given;
mass of the ball, m = 0.175 kg
initial displacement of the ball, h₁ = 1.25 m
final displacement of the ball, h₂ = 0.805 m
<u>Assumptions:</u>
let the downward direction of the ball be positive
let the upward direction of the ball be negative
The following equation of motion will be used to determine the final velocity of the ball at each displacement.
v² = u² ± 2gh
The final velocity of the ball when it is dropped downwards to 1.25 m;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 1.25)
v = 4.95 m/s
The final velocity of the ball when it rebounds from the floor to 0.805 m;
vf² = u² - 2gh
vf² = 0² - 2gh
vf² = -2gh
vf = -√2gh
vf = - √(2 x 9.8 x 0.805)
vf = -3.97 m/s
The impulse received by the ball is calculated as;
J = ΔP = mΔv = m(vf - v)
= 0.175(-3.97 - 4.95)
= - 1.561 kg.m/s
The negative sign indicates upward direction of the impulse.