Answer:
The length of rod A will be <u>greater than </u>the length of rod B
Explanation:
We, know that the formula for final length in linear thermal expansion of a rod is:
L' = L(1 + ∝ΔT)
where,
L' = Final Length
L = Initial Length
∝ = Co-efficient of linear expansion
ΔT = Change in temperature
Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.
For Rod A:
∝₁ = 12 x 10⁻⁶ °C⁻¹
For Rod B:
∝₂ = β₂/3
where,
β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹
Therefore,
∝₂ = 24 x 10⁻⁶ °C⁻¹/3
∝₂ = 8 x 10⁻⁶ °C⁻¹
Since,
∝₁ > ∝₂
Therefore,
L₁ > L₂
So, the length of rod A will be <u>greater than </u>the length of rod B
The two fields were physical quantities are used in motion calculations are length and mass with time.
The physical quantity in a field is referred as every point in a particular space time.
<h3>
How physical quantities are used in motion calculations?</h3>
If we consider an object, the physical property of the object is considered as physical quantity and to measure that object is known as units. The Physical quantity can be classified as elemental physical quantity and derived physical quantity. Length, mass, time, etc.. are elemental physical quantity, momentum, density, acceleration, etc... are derived physical quantity. Only for charge and temperature the physical quantity will be less than zero.
Length, mass and time are the physical quantities used in motion calculations.
Learn more about motion calculations,
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Dark matter may explain <span>unexpected orbital velocities of stars in galaxies.</span>
Answer:
4.02 s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 35°
Initial velocity (u) = 50 m/s
Acceleration due to gravity (g) = 10 m/s²
Time of flight (T) =?
The time of flight of the arrow can be obtained as follow:
T = 2uSineθ / g
T = 2 × 35 × Sine 35 / 10
T = 70 × 0.5736 / 10
T = 7 × 0.5736
T = 4.02 s
Therefore, the time taken for the arrow to return is 4.02 s
