x-component is V·cos(angle)
y-component is V·sin(angle)
When the angle is 45°, its sin and cos are both 1/2·√2 .
Vx = 20·√2 = 28.28 m/s
Vy = 20·√2 = 28.28 m/s
What color is a carrot?
Answer:
reddish-orrange
Explanation:
please mark me as brainliest
A uniform 140 g rod with length 57 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 30 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 11 cm on each sides of the center, at which time the system rotates at an angular speed of 23 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
Answer:
The correct answer is "12 rad/s"
Explanation:
The given values are,
Mass of rod,
M = 140 g
i.e.,
= 0.14 kg
Length,
L = 57 cm
i.e.,
= 0.57 m
Mass of beads,
M = 30 g
i.e.,
= 0.03 kg
Angular speed,
r = 11 cm
i.e.,
= 0.11 m
Now,
The inertia of rods will be:
= [tex]\frac{1}{12}ML ^2[/tex]
On substituting the values, we get
= [tex]\frac{1}{12}\times 0.14\times (0.57)^2[/tex]
= [tex]0.0037905 \ kg-m^2[/tex]
The inertia of beads will be:
= [tex]mr^2[/tex]
On substituting the values, we get
= [tex]0.03\times (0.11)^2[/tex]
= [tex]0.000726 \ kg-m^2[/tex]
The total inertia will be:
= [tex]Inertia \ of \ rods+Inertia \ of \ beads[/tex]
= [tex]0.0037905 + 0.000726[/tex]
= [tex]0.0045165 \ kg-m^2[/tex]
now,
The angular speed of the system will be:
⇒ [tex]L_1w_1=L_2w_2[/tex]
On substituting the values in the above equation, we get
⇒ [tex]0.0045165\times 23 = (0.0037905 + (0.03\times 0.285^2)\times 2 )\times w_2[/tex]
⇒ [tex]0.1038795 = 0.0037905 + (0.00243675\times 2 )\times w_2[/tex]
⇒ [tex]w_2 = 12 \ rad/s[/tex]
ball A is dropped from a hot air balloon rising at a costant velocity of 14,7 m.s'1 at a height of 19,7 m above the ground.the ball took 1.5s to reach its maximum height and hits the ground after some time in air.ignore the effects of air resistanceUse the ground as zero reference.3.1.1calculate the maximum height reached by the ball above the ground
Answer:
this slow site thinks the answer is a link
Explanation:
this was a week ago so i dont know if u still need help
What is the magnitude of the resultant vector? Round
your answer to the nearest tenth.
m
R
5 m
13 m
Intro
Done
Answer: 13.9 m
Explanation:
Answer:
13.9m
Explanation:
Answer on Edge
An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you can plz
Answer:
P = 11666.6 W
Explanation:
Given that,
Work done by the motor, W = 3500 kJ
Time, t = 5 min = 300 s
We need to find the power developed by the motor. Power developed is given by :
[tex]P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W[/tex]
So, the required power is 11666.6 W.
A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80 kg block that is at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3 m/s. What is the speed of the ball just after the collision
Answer:
the speed of the ball just after the collision is 1.5 m/s.
Explanation:
Given;
mass of the ball, m₁ = 1.6 kg
initial velocity of the ball, u₁ = 0
mass of the block, m₂ = 0.8 kg
initial velocity of the block, u₂ = 0
final velocity of the block, v₂ = 3 m/s
let the final velocity of the ball after collision = v₁
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
1.6 x 0 + 0.8 x 0 = 1.6 x v₁ + 0.8 x 3
0 = 1.6v₁ + 2.4
-1.6v₁ = 2.4
v₁ = -2.4 / 1.6
v₁ = - 1.5 m/s
v₁ = 1.5 m/s (in opposite direction of the block)
Therefore, the speed of the ball just after the collision is 1.5 m/s.