Answer:
Find attached of the diagram that complete the question.
Velocity = 1.5 sec²θ m/s
Acceleration = 9 sec²θtanθ m/s²
Explanation:
Calculating the velocity:
V = dx/dt
x = 0.5tanθ
θ = 3t
ö = 3
V = d(0.5tanθ)/dt
= 0.5*sec²θ*ö
= 3*0.5sec²θ
= 1.5 sec²θ m/s
Calculating the acceleration, we have;
a = dv/dt
= d(1.5sec²θ)/dt
= 1.5*2sec²θtanθ*ö
= 1.5*2sec²θtanθ*3
= 9sec²θtanθ m/s²
The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.
Answer:
False
Explanation:
In electric heater electric energy is converted into heat energy. In heater wires are present which have resistance and current is flow in heater when we connect the heater to supply.
And we know that whenever current is flow in any resistance then heat is produced so in electric heaters electric energy is converted into heat energy
So this is a false statement
Answer:
The external pressure is p = -21.9 psf or p = -8.85 psf
Explanation:
Given :
Velocity of wind, v = 120 mi / hr
(wind direction factor)
= topographical factor (for flat terrain)
= velocity pressure at height h
But for height h = 30 ft, 
= 0.98 (from table)
= 36.16
Now, 
, so 
(from table)
where, p = external pressure
G = 0.85 = gust factor (for typical rigid building)
(internal pressure co efficient)
Therefore putting the values,
p = -21.9 psf or p = -8.85 psf
Answer:
The coefficient of linear expansion of the metal is ∝ = 2.91 x 10⁻⁵ °C⁻¹.
Explanation:
We know that Linear thermal expansion is represented by the following equation
Δ L = L x ∝ x Δ T ---- (1)
where Δ L is the change in length, L is for length, ∝ is the coefficient of linear expression and Δ T is the change in temperature.
Given that:
L = 0.6 m
T₁ = 15° C
T₂ = 37° C
Δ L = 0.28 mm
∝ = ?
Solution:
We know that Δ T = T₂ ₋ T₁
Putting the values of T₁ and T₂ in above equation, we get
Δ T = 37 - 15
Δ T = 22 °C
Also Δ L = 0.28 mm
Converting the mm to m
Δ L = 0.00028 m
Putting the values of Δ T, Δ L, L in equation 1, we get
0.00028 = 0.6 x ∝ x 22
Rearranging the equation, we get
∝ = 0.00028 / (0.6 x 16)
∝ = 0.00028 / 13.2
∝ = 2.12 x 10⁻⁵ °C⁻¹
