Answer:
Explanation:
That an optical illusion somehow interferes with the way we see things. Even simple illusions can completely fool us. If you search out the term, you'll see all kinds of them.
Most critically we see one thing and know another to be true. But knowing the truth doesn't help us. We still see and believe the truth of the illusion.
Answer:
v = 98.75 km/h
Explanation:
Given,
The distance driver travels towards the east, d₁ = 135 km
The time period of the travel, t₁ = 1.5 h
The halting time, tₓ = 46 minutes
The distance driver travels towards the east, d₂ = 215 km
The time period of the travel, t₁ = 2 h
The average speed of the vehicle before stopping
v₁ = d₁/t₁
= 135/1.5
= 90 km/h
The average speed of vehicle after stopping
v₂ = d₂/t₂
= 215/2
= 107.5 km/h
The total average velocity of the driver
v = (v₁ +v₂) /2
= (90 + 107.5)/2
= 98.75 km/h
Hence, the average velocity of the driver, v = 98.75 km/h
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
The answer is C.
The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
