Answer:
0.06 N
1.08 m/s
Explanation:
m = mass of the fan cart = 0.250 kg
a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²
F = Net force on the cart
Net force on the cart is given as
F = ma
F = (0.250) (0.24)
F = 0.06 N
v₀ = initial velocity of the cart = 0 m/s
v = final velocity of the cart
t = time interval = 4.5 s
Using the equation
v = v₀ + a t
v = 0 + (0.24) (4.5)
v = 1.08 m/s
When a crest-trough meet the interference produced will be destructive in nature hence they both will cancel out and the amplitude produced will be equal to zero hence the loudness will reduce to zero.
Answer:
0.558 atm
Explanation:
We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT
Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:
P₀ = P₁ + P₂ + ....
P₀= total pressure
P₁=P₂= is the partial pressure of each gass
If we can consider that each gas is an ideal gas, then:
P₀= (nRT/V)₁ + (nRT/V)₂ +..
Considering the molecular mass of O₂:
M O₂= 32 g/mol
And also:
R= ideal gas constant= 0.082 Lt*atm/K*mol
T= 65°C=338 K
4.98 g O₂ = 0.156 moles O₂
V= 7.75 Lt
Then:
P°O₂=partial pressure of oxygen gas= (0.156x0.082x338)/7.75
P°O₂= 0.558 atm
Well, they're not quite the way Newton expressed it, but out of all this mess of statements, there are two that are correct AND come from Newton's 2nd Law of Motion:
<em>-- The smaller the mass of an object, the greater the acceleration of that object when a force is applied. </em>
<em>-- The greater the force applied, the greater the acceleration.</em>
For the <u><em>other </em></u>statements in the question:
-- <em>Every reaction is equal to the force applied.</em> True; comes from Newton's <u><em>3rd</em></u> law of motion.
-- <em>Forces are balanced when they are equal and opposite.</em> True; kind of a definition, not from Newton's laws of motion.
-- <em>An object at rest or in motion will remain at rest or in motion unless acted upon by an unbalanced force.
</em> True; comes from Newton's <em><u>1st </u></em>law of motion.
Answer:
Change in Q = 2.1x 10^-3 C
Explanation:
We are given that
The Initialcapacitance C1 = 6.0μF
Initial charge oncapacitor
Q1 = C1 V
= 6.00 x 10^-6 x 100
= 6.00 x 10^-4 C
So the Final capacitance C2 will be
= K x C1 = 4.5 x 6.00 x 10^-6
= 2.7 x 10^ -5 F
So to get Finalcharge
We use Q2 = C2 x V
= 2.7 x 10^ - 5 x 100
= 27 x 10^ -4 C
So Charge flown in thecapacitor is change in Q
Which is = Q2 - Q1
= 27 x 10^-4 - 6.0 x 10^ -4
Change in Q = 2.1x 10^-3 C
