Answer:
1. bending of light in gravitational fields.
2. effect of gravitational redshift.
3. perihelion precission of mecury.
Explanation:
1 bending of light in gravitational fields, we can think of it like this:
by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.
2. effects of gravitational redshift:
this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.
3. perihelion precission of mecury:
according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.
Answer:
V = 4.63 m/s
V = 11.31 m/s
Explanation:
Given,
The distance traveled by the bus, towards north, d = 2.5 km
= 2500 m
The time taken by the trip is, t = 9 min
= 540 s
The velocity of the bus,
V = d / t
= 2500 / 540
= 4.63 m/s
At another point, the bus travels at a constant speed of v = 18 m/s
Therefore the velocity becomes
V = (4.63 + 18)/2
= 11.31 m/s
Hence, the velocity of the bus, V = 11.31 m/s
Answer:
0.45 seconds
Explanation:
Letting the value of g = 10 m/s/s
final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)
initial velocity(u) = 4.5 m/s
acceleration = -10 m/s/s (since the gravity is acting against the egg)
time = t seconds
From the first equation of motion:
<em>v = u + at</em>
<em>0 = 4.5 + (-10)t</em>
<em>t = -4.5 / -10</em>
t = 0.45 seconds
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
