When the mass of the spring changed from 0.2kg to 0.1kg, the time period changed from 1 sec to 0.5 seconds
<u>Explanation:</u>
Given-
Mass, m1 = 0.2kg
Time period, T1 = 1s
m2 = 0.1 kg
T2 = ?
We know,
where,
T = Time period
m = mass
k = spring constant
From the equation, we can see that T is directly proportion to the square root of mass, m
T ∝ √m
So,
If m1 = 0.2kg , T1 = 1s and m2 = 0.1kg
The T2 would be:
Therefore, when the mass of the spring changed from 0.2kg to 0.1kg, the time period changed from 1 sec to 0.5 seconds
when you jump up in Gravity it pulls you down the less gravity the more you can jump easier
Answer:
1.29 moles
0.753 moles
0.745 moles
Explanation:
PV=nRT
n=PV/RT
n=(1)(34.2)/(0.0821)(323.7)
n=1.29
n=PV/RT
n=(1)(22.4)=(0.0821)(362.15)
n=0.753
n=PV/RT
n=(1)(16.7)/(0.0821)(273.15)
n=0.745
In the ideal gas equation, T is measured in Kelvin.
Answer:
1)31/3Ω
2)18/31A
3)4.06V
Explanation:
According to the diagram and 10 Ω resistors are parallel to each other, parallel resistor can be calculated using the formula below
1/R= 1/R1 + 1/R2
But we know R1= 5 Ω and R2= 10 Ω
1/R= 1/5 + 1/10
R= 10/3 Ω
=3.33 Ω
Then if we follow the given figure, 10/3 Ω and 7Ω are now in series then
Req = 10/3 +7
Req= 31/3 Ω
Therefore, equivalent resistance = 31/3 Ω
According to ohms law we know that V= IR
Then I= V/R
Where I= current
R= resistance
V= voltage
I= 6/(31/3)
I= 18/31A
We can now calculate the voltage accross the resistor which is
V=(18/31)× 7
V=4.06V
Therefore, the voltage accross the 7 ohm resistor is 4.06V
CHECK THE FIQURE AT THE ATTACHMENT
