Answer:
R (120) = 940Ω
Explanation:
The variation in resistance with temperature is linear in metals
ΔR (T) = R₀ α ΔT
where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC
let's calculate
ΔR = 1000 (-0,0005) (120-0)
ΔR = -60
Ω
ΔR = R (120) + R (0) = -60
R (120) = -60 + R (0)
R (120) = -60 + 1000
R (120) = 940Ω
Answer:
4.37 * 10^-4 J
Explanation:
Energy stored :
mgΔl / 2
m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m
Δl = mgl / πr²Y
Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m
Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10
Δl = 98 / 3.5 * π * 10^6
Δl = 0.00000891267
Energy stored :
mgΔl / 2
(10 * 9.8 * 0.00000891267) / 2
= 0.00043672083 J
4.37 * 10^-4 J
Answer:
96 m
Explanation:
Given,
Initial velocity ( u ) = 4 m/s
Final velocity ( v ) = 20 m/s
Time ( t ) = 8 s
Let Acceleration be " a ".
Formula : -
a = ( v - u ) / t
a = ( 20 - 4 ) / 8
= 16 / 8
a = 2 m/s²
Let displacement be " s ".
Formula : -
s = ut + at² / 2
s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2
= 32 + ( 2 ) ( 64 ) / 2
= 32 + ( 2 ) ( 32 )
= 32 + 64
s = 96 m
Therefore, it travels 96 m in time 8 s.
<h2>
a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b)
Factor of increase in weight is 27.95</h2>
Explanation:
a) Acceleration due to gravity
Here we need to find acceleration due to gravity of Sun,
G = 6.67259 x 10⁻¹¹ N m²/kg²
Mass of sun, M = 1.989 × 10³⁰ kg
Radius of sun, r = 6.957 x 10⁸ m
Substituting,
Acceleration due to gravity on the surface of the Sun = 274.21 m/s²
b) Acceleration due to gravity in earth = 9.81 m/s²
Ratio of gravity = 274.21/9.81 = 27.95
Weight = mg
Factor of increase in weight = 27.95
Answer:
The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Explanation:
P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.
Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.
Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
