Answer: acceleration = -10m/s^2
Explanation:
Use Fab 5 equation to solve for acceleration. (Use equation without d/distance because it is irrelevant for this particular question)
vi = 60m/s
vf = 0m/s because car comes to a stop
t = 6s
a = ?
d = irrelevant
Use a = vf-vi/t
Step by step solution:
a = vf - vi / t
a = 0 - 60 / 6
a = -60 / 6
a = -10m/s^2
Hope this helps! :)
Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Explanation:
Given that;
diameter of the mirror d = 1.7 m
height h = 180 km = 180 × 10³ m
wavelength λ = 500 nm = 5 × 10⁻⁹ m
Now Angular separation from the peak of the central maximum is expressed as;
sin∅= 1.22 λ / d
sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7
sin∅ = 3.588 × 10⁻⁷
we know that;
sin∅ = object separation / distance from telescope
object separation =
sin∅ × distance from telescope
object separation = 3.588 × 10⁻⁷ × 180 × 10³
object separation =6.45 × 10⁻² m
then we convert to centimeter
object separation = 6.45 cm
Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Answer:
Explanation:
Young's modulus of elasticity Y = stress / strain
stress = force / cross sectional area
= weight of 15 kg / π r²
= 15 x 9.8 / 3.14 x ( .025 x 10⁻² )²
stress = 74.9 x 10⁷ N / m²
strain = Δ L / L , Δ L is change in length and L is original length
Putting the values
strain = .0168 / 2.7 =.006222
Young's modulus of elasticity Y = 74.9 x 10⁷ / .006222
= 120.88 x 10⁹ N / m² .
Answer:
The perceived frequency is higher than the actual emitted sound frequency. that means that the received sound waves are of shorter wavelength.
Explanation:
When the source of a sound wave is moving toward the observer, the perceived frequency of the wave changes in relation to the observer producing a change in pitch. The effect is called Doppler effect in honor of the physicist who formulated the physical explanation.
In the case of the sound source approaching the observer, the perceived frequency is higher than that actually emitted by the sound source.
