Answer:
V = 20 miles /sec
Explanation:
We have remaining distance = d = 96 miles
Lets call Pascal velocity V in miles per hour
Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t
1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁
And in the case of reducing his velocity
(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384
So we a 2 equation system with two uknown variables
1.5*V = 96/t₁ (1)
V*t₁ + 16*V = 384 (2)
We solve from equation (1) t₁ = 64/V
And by substitution in equation (2)
V * (64/V) + 16* V = 384
64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16
V = 20 miles /sec
The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
Typically no. Displacement can be in multiple directions as a vector. of something is traveling only along x, then it would be true though this is usually not the case.
Answer:
The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.
