Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet
Answer:
car B will be 30 Km ahead of car A.
Explanation:
We'll begin by calculating the distance travelled by each car. This is illustrated below:
For car A:
Speed = 40 km/h
Time = 3 hours
Distance =?
Speed = distance / time
40 = distance / 3
Cross multiply
Distance = 40 × 3
Distance = 120 Km
For car B:
Speed = 50 km/h
Time = 3 hours
Distance =?
Speed = distance / time
50 = distance / 3
Cross multiply
Distance = 50 × 3
Distance = 150 Km
Finally, we shall determine the distance between car B an car A. This can be obtained as follow:
Distance travelled by car B (D₆) = 150 Km
Distance travelled by car A (Dₐ) = 120 Km
Distance apart =?
Distance apart = D₆ – Dₐ
Distance apart = 150 – 120
Distance apart = 30 Km
Therefore, car B will be 30 Km ahead of car A.
Answer:
Charge of particle 2, 
Explanation:
Given that,
Charge 1, 
The distance between charges, r = 0.241 m
Force experienced by particle 1, F = 3.44 N
We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :
or
So, the magnitude of electric charge 2 is . Since, the force is attractive then the magnitude of charge 2 must be negative.
C. making fun of a peer because she is Asian
hope this helps
Density = (mass) / (volume)
= (48 g) / (6 cm³)
= (48 / 6) (g / cm³)
= 8 g/cm³
