Making a drawing of the system, we will have two forces which are tension and the weight of the object. Balancing the forces present, we do as follows:
T = W
W = 30 N
Therefore, weight is equal to 30 N. Hope this answers the question. Have a nice day. Feel free to ask more questions.
Answer:
R = 4Ω
Explanation:
If we have two resistors with resistances R1 and R2 in series the total resistance is R = R1 + R2
If the resistances are in parallel, the total resistance is given by:
1/R = 1/R1 + 1/R2.
First, we have a resistor with R1 = 1.5Ω
This resistor is connected in series with a parallel part (let's find the resistance of this parallel part), in one branch we have two resistors in series with resistances:
R2 = 8Ω and R3 = 4Ω
Because these are in series, the resistance of that branch is:
R = 8Ω + 4Ω = 12Ω
In the other branch, we have a single resistor of R4 = 4Ω
The resistance of the parallel part is:
1/R = 1/12Ω + 1/4Ω = 1/12Ω + 3/12Ω = 4/12Ω = 1/3Ω
1/R = 1/3Ω
R = 3Ω
Then we have a resistor (the first one, R1 = 1.5Ω) in series with a resistor of 3Ω.
Then the total resistance is:
R = 1Ω + 3Ω = 4Ω
coefficient of static friction is 1.7329
Explanation:
given data
velocity = 60 mph
acceleration = 17 m/s²
to find out
coefficient of static friction
solution
we will apply here centripetal force equation
that is
m×v²/r = µ × m × g ..................1
here v²/r is centripetal acceleration and m is mass and µ is coefficient static friction so
µ = a / g
µ = 17 / 9.81
µ = 1.7329
so coefficient of static friction is 1.7329
Answer:
The centre of sphere of part of which a lens is formed is called the centre of curvature of the lens.
Answer:
Explanation:
Lead has an half life of 43.5 seconds
The radioactive decay law says
N(t) = No•2^(-t / t½)
Hence, the percent of a sample of the Lead-214 isotope that decay after a time equivalent to 7 half-lives
N(t) = No•2^(-t / t½)
N(t) / No = 2^(-t / t½)
N(t) / No = 2^-7
N(t) / No = 7.81 × 10^-3
To percentage
N(t) / No = 7.81 × 10^-3 × 100
N(t) / No = 0.781 %