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.
The answer is true...............
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
A type O star is likely to appear blue.
Answer:
Explanation:
When the spring is compressed by .80 m , restoring force by spring on block
= 130 x .80
= 104 N , acting away from wall
External force = 82 N , acting towards wall
Force of friction acting towards wall = μmg
= .4 x 4 x 9.8
= 15.68 N
Net force away from wall
= 104 -15.68 - 82
= 6.32 N
Acceleration
= 6.32 / 4
= 1.58 m / s²
It will be away from wall
Energy released by compressed spring = 1/2 k x²
= .5 x 130 x .8²
= 41.6 J
Energy lost in friction
= μmg x .8
= .4 x 4 x 9.8 x .8
= 12.544 J
Energy available to block
= 41.6 - 12.544 J
= 29 J
Kinetic energy of block = 29
1/2 x 4 x v² = 29
v = 3.8 m / s
This will b speed of block as soon as spring relaxes. (x = 0 )
