Answer:
a galaxy observed at a distance of 5 billion light-years
Answer:
W = 34.64 ft-lbs
Explanation:
given,
Horizontal force = 4 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
W = F.d cos θ
W = 4 x 10 x cos 30°
W = 40 x 0.8660
W = 34.64 ft-lbs
Hence, work done on the box is equal to W = 34.64 ft-lbs
Answer:
200 N
Explanation:
Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.
So, E = σ/ε = FL/eA
Now, since at break extension = e.
So making e subject of the formula, we have
e = FL/EA = FL/Eπr² where r = radius of metal wire
Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r
So, e' = F'(2L)/Eπ(2r)²
e' = 2F'L/4Eπr²
e' = F'L/2Eπr²
Since at breakage, both extensions are the same, e = e'
So, FL/Eπr² = F'L/2Eπr²
F = F'/2
F' = 2F
Since F = 100 N,
F' = 2 × 100 N = 200 N
So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
