Answer:
1200 m
Explanation:
s= vt
s= 20m/s × 60 s
s= 1200 m
hope it hepls
for any question comment me ❤❤
Https://www.ted.com/topics/biotech
They made me do it I don’t even know what to say I’m so sorry
Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00

At x = 1.00 m

= 4J
Kinetic energy = (1/2)mv²

= 18J
Total energy will be =
4J + 18J = 22J
At x = 5

= -0.24J
Kinetic energy =

= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2

= 3.33 m/s
b) magnitude of force when x = 5.0m



At x = 5.0 m


= 0.016N
Answer:
0.776 m far Pinhole should be placed before the viewing screen
Explanation:
For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

where
is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm
Pinhole should be placed before the viewing screen is
