Answer:
(a) 152.85 Nm
(b) 1528.5 Nm
Explanation:
According to the formula of power
P = τ ω
ω = 2 π f
(a) f = 2500 rpm = 2500 / 60 = 41.67 rps
So, 40 x 1000 = τ x 2 x 3.14 x 41.67
τ = 152.85 Nm
(b) f = 250 rpm = 250 / 60 = 4.167 rps
So, 40 x 1000 = τ x 2 x 3.14 x 4.167
τ = 1528.5 Nm
Answer:
The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s
Explanation:
Given,
Your velocity, V₁ = 2 m/
The velocity of the person, V₂ =?
The velocity of the person relative to you, V₂₁ = 3 m/s
According to the relative velocity of two
V₂₁ = V₂ -V₁
∴ V₂ = V₂₁ + V₁
On substitution
V₂ = 3 + 2
= 5 m/s
Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s
Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up
0N. The net force acting on this firework is 0.
The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.
Based on the free-body diagram, we have:
The force of gases is Fgases = 9,452N
The force of the rocket Frocket = -9452
Then, the net force acting is:
Fnet = Fgases + Frocket
Fnet = 9,452N - 9,452N = 0N
Answer: Relative motion
Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.
The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p
Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.
Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.
Motion of this nature is known as relative motion.
<em>Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid</em>
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For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.
The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.
