Answer:
Option 2. √3 x V
Explanation:
Let the velocity of satellite orbiting Earth at r be V
Let the velocity of the satellite orbiting at 3r be V1
V (orbiting at r) = √(2gr)
V1 (orbiting at 3r) = √(2g3r)
Now let us find the ratio of V1(orbiting at 3r) to V(orbiting at r) .
This is illustrated below
V1 / V = √(2g3r)/√(2gr)
V1 / V = √3
Cross multiply to express in linear form
V1 = V x√3
V1 = √3 x V
From the above illustrations, we can see that the velocity of the satellite when it is moved to an orbital radius of 3r is: √3 x V
Answer:
Explanation:
The boy throw the pencil upward at a speed of 6.33 m/s
Then,
Initial velocity of throw is 6.33 m/s
u = 6.33 m/s.
Time to reach a maximum height of 1.25m
h = 1.25m
Note: at maximum height, the final velocity is zero
v = 0m/s
Acceleration due to gravity is
g = 9.81m/s²
We want to calculate time to reach maximum height
t = ?
Then, applying equation of motion
v = u + gt
But since it is against gravity, then, g is negaive
Then,
v = u - gt
0 = 6.33 - 9.81t
-6.33 = -9.81t
Then,
t = -6.33 / -9.81
t = 0.645 seconds
Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
Answer:
44.6 N
Explanation:
Draw a free body diagram of the block. There are four forces on the block:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force F pulling right 30° above horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30° − mg = 0
N = mg − F sin 30°
Sum of forces in the x direction:
∑F = ma
F cos 30° − Nμ = 0
F cos 30° = Nμ
N = F cos 30° / μ
Substitute:
mg − F sin 30° = F cos 30° / μ
mg = F sin 30° + (F cos 30° / μ)
Plug in values:
mg = 20 N sin 30° + (20 N cos 30° / 0.5)
mg = 44.6 N
Answer: A. Her speed is 4.4 m/s, and her velocity is 0 m/s.
Explanation: i took the test on edgenuity
