Answer:
100.11 m
Explanation:
Given:
Coefficient of volume expansion, 
/°C.
Initial length of rod, 
Change in temperature, ΔT=
°C
We know that,
/°C
Here, 
is the coefficient of linear expansion.
Now, we know that,
Change in length (Δl) is given as,
Plug in all the values and solve for 
.
This gives,
Therefore, the length of rod after expansion is 
m.
Answer:
Given:
radius of the coil, R = 6 cm = 0.06 m
current in the coil, I = 2.65 A
Magnetic field at the center, B = 
Solution:
To find the number of turns, N, we use the given formula:
Therefore,
N = 22.74 = 23 turns (approx)
By analyzing the vector, the clothesline tension is 25 N.
We need to know about vectors to solve this problem. Force is included in vectors that have magnitude and direction. It can be written as
F = (Fx i + Fy j) N
where F is the force vector, Fx is the x-axis component and Fy is the y-axis component.
From the question we know that :
x = 10 m
(midpoint = 5m)
y = 1 m
m = 1 kg
Find the weight
W = m . g
W = 1 x 10
W = 10N
There are two y-axis components of the tension that hold the clothes. Because of static conditions, we can write
∑Fy = 0
Ty + Ty - W = 0
2Ty - 10 = 0
2Tsinθ - 10 = 0
Find sinθ
sinθ = y/x
sinθ = 1/5
Substitute the sinθ
2Tsinθ - 10 = 0
2T(1/5) - 10 = 0
2T/5 = 10
T = 25 N
Hence, the clothesline tension is 25 N.
For more about vectors at: brainly.com/question/25811261
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Answer:
W ’= 21.78 kg
Explanation:
The expression for weight is
W = m g
let's look for the acceleration of gravity with the universal law of gravitation
F = G m M / r2
F = m (G M / r2)
without comparing the two equations
g’= G M / r2
in that case M = 2 Mo and r = 3 ro
where mo and ro are the mass and radius of the earth
we substitute
g ’= G 2Mo / (3r₀) 2
G ’= 2/9 G Mo / r₀²
g ’= 2/9 g
the weight of the body on this planet is
W ’= m g’
W ’= m 2/9 g
let's calculate
W ’= 2/9 10 9.8
W ’= 21.78 kg
