Answer:
ALMA is a trans-formative radio telescope that can study cosmic light that straddles the boundary between radio and infrared. Most objects in the Universe emit this kind of energy, so the ability to detect it has been a driver for astronomers for decades.
Explanation:
Answer:
3.88 × 10^-4 m
Explanation:
Given that a person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
The total weight = 614 + 85
The total weight = 699N
Let the total area of contact = A
Pressure = Force / A
Substitute all the parameters into the formula
900000 = 699 /A
A = 699 / 900000
A = 7.77 × 10^-4 m
The area of contact between each tire and the ground will be = A/2
That is, 7.77 / 2 = 3.88 × 10^-4 m
Therefore, the area of contact between each tire and the ground is 3.88 × 10^-4 m approximately.
Answer:
Incomplete question: "A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?"
The normalized corrected value at 32.5 mm is 0.1638
Explanation:
The corrected light measurement at 29 mm is equal to:
20.7 - 5.1 = 15.6 mV
The corrected light measurement at 32.5 mm is equal to:
15.6 - 5.1 = 10.5 mV
To normalize the data to its maximum value means that the maximum value must be calculated and the data must be scaled using that value, as in this case the maximum value is 15.6 mm, then the normalized corrected value at 32.5 mm is equal to:
10.5 * 15.6 = 163.8 = 0.1638
Answer:
The formula for density is d=M/V
Explanation:
where d is density, M id mass, V is volume.
Answer:
The displacement, velocity and time is 20.4 m, -4.52 m/s and 3.624 sec.
Explanation:
Given that,
Initial velocity =15.0 m/s
Height = 10 m
We need to calculate the displacement at 2 sec
Using equation of motion

Put the value into the formula

At 2 sec,


We need to calculate the time
Using equation of motion again

At ground s = 0,



Neglect the negative term because the time is not negative
We need to calculate the velocity
Using formula of velocity



Hence, The displacement, velocity and time is 20.4 m, -4.52 m/s and 3.624 sec.