<span>A plane flies along a straight line path after taking off, and it ends up 210 km farther east and 90.0 km farther north, relative to where it started. we should use Pythagoras theorem for this.
distance^2= 210^2 + 90^2.=228.47km, when rounding off this would comes around 230 km</span>
Answer:
(a) 47.08°
(b) 47.50°
Explanation:
Angle of incidence = 78.9°
<u>For blue light :
</u>
Using Snell's law as:
Where,
Θ₁ is the angle of incidence
Θ₂ is the angle of refraction
n₂ is the refractive index for blue light which is 1.340
n₁ is the refractive index of air which is 1
So,
Angle of refraction for blue light = sin⁻¹ 0.7323 = 47.08°.
<u>For red light :
</u>
Using Snell's law as:
Where,
Θ₁ is the angle of incidence
Θ₂ is the angle of refraction
n₂ is the refractive index for red light which is 1.331
n₁ is the refractive index of air which is 1
So,
Angle of refraction for red light = sin⁻¹ 0.7373 = 47.50°.
So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2.
Answer:
a) 6.26 m/s
b) 7.67 m/s
Explanation:
The potential energy at height h0 is initially ...
PE0 = mgh0
At height h1, the potential energy is ...
PE1 = mgh1
The difference in potential energy is converted to kinetic energy:
PE0 -PE1 = KE1 = (1/2)m(v1)^2
Solving for v1, we have ...
mg(h0 -h1) = (1/2)m(v1)^2
2g(h0 -h1) = (v1)^2
v1 = √(2g(h0 -h1))
__
a) When the body is 1 m high, its speed is ...
v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high
__
b) When the body is 0 m high, its speed is ...
v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground
3. An object will float if its density is less than the matter it is submerged in.
If it sinks in water, its density is more than 1. If it floats in glycerin, its density is less than 1.26.
