Answer:
10.66mm
Explanation:
Given that the elongation is and Length, and Tensile Strength,, The longitudinal stress can be calculated as:
The value of strain-stress, corresponding to '
Hence, the radius of the specimen is 10.66mm
In 1 year, 365 days, each day 24 hours, each hour 60 minutes and each minute 60 seconds.
OR
.
As the radius of the earth's very nearly circular orbit around the sun is 1.5 x 10^11 m.
Therefore, the magnitude of the earth's velocity
.
The magnitude of the earth's angular velocity is
.
The magnitude of the earth's centripetal acceleration is
Answer:
x_total = 0.17m
Explanation:
We can treat this exercise with the kinematics equations, where in the first part it is accelerated and in the second it is a uniform movement.
Let's analyze accelerated motion
The time that lasts is t = 20 10⁻³ s, the initial speed is zero (v₀ = 0), let's find the length that advances
x₁ = v₀ t + ½ a t²
x₁ = ½ a t²
x₁ = ½ 210 (20 10⁻³)²
x₁ = 4.2 10⁻² m
let's find the speed for the end of this movement
v = v₀ + a t
v = 0 + 210 20 10⁻³
v = 4.2 m / s
with this speed we can find the distance that the uniform movement
x₂ = v t2
x₂ = 4.2 30 10⁻³
x₂ = 1.26 10⁻¹ m
x₂ = 0.126m
the total distance traveled is
x_total = x₁ + x₂
x_total = 0.0420 +0.126
x_total = 0.168m
Let's reduce the significant figures to two
x_total = 0.17m
Average Acceleration = (change in speed) / (time for the change)
Change in speed = (ending speed) - (starting speed)
= 15 m/s - 24 m/s = -9 m/s
Acceleration = (-9 m/s) / (12 sec) = - 0.75 m/s² .
If the 30 N on the rope were pulled straight up, it would offset the force of gravity ( m g = 10 kg * 9.8 N/kg = 98 N) , leaving a net force up from the ground on the sled of 98-30 = 68 N. Since the rope is pulled at the angle of 25o, only part of the force is in the upward direction, (30N)(sin(25) = (30)(.423) = 12.7. So the net force becomes 98 N down offset by 12.7 up or 98-12.7 = 85.3 N. Ah, there it is: C.