Answer:
As force = tension in string so tension in string is 400N
Explanation:
Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²
Answer:0.38
Explanation:
the formula is f = c / λ
so f= 2.5/6.5
and that equals 0.38 46 and so on so i just rounded it
Explanation:
Use the height of the cliff to determine how long it took the car to land.
Take down to be positive. Given:
Δy = 7.93 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
7.93 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.27 s
Use the time to calculate the horizontal velocity.
Given:
Δx = 26.7 m
a = 0 m/s²
Find: v₀
Δx = v₀ t + ½ at²
26.7 m = v₀ (1.27 s) + ½ (0 m/s²) (1.27 s)²
v₀ = 21.0 m/s
The driver was going 21.0 m/s, faster than the speed limit of 9.72 m/s.
