Answer: acceleration a = 25m/s^2
Explanation:
Given that:
The plane travels with constant acceleration
x1 = 241.22 m at t1 = 3.70 s
x2 = 297.60 m at t2 = 4.20 s
x3 = 360.23 m at t3 = 4.70 s.
We need to calculate the velocity in the two time intervals.
Interval 1:
Average Velocity v1 = ∆x/∆t = (x2 - x1)/(t2-t1)
v1 = (297.60-241.22)/(4.20-3.70) = 112.76m/s
Interval 2:
Average Velocity v2 = ∆x/∆t = (x3-x2)/(t3-t2)
v2 = (360.23-297.60)/(4.70-4.20)
v2 = 125.26m/s
Acceleration:
Acceleration a = ∆v/∆t
∆v = v2-v1 = 125.26m/s-112.76m/s = 12.5m/s
∆t = change in average time of the two intervals = (t3-t1)/2 = (4.70-3.70)/2 = 0.5s
a = 12.5/0.5 = 25m/s^2
Answer:
a = -8.912 m/s²
Explanation:
Given,
The initial velocity of the car, u = 28 m/s
The final velocity of the car, v = 0
The distance traveled by car, d = 88 m
The velocity displacement relation is given by the formula
v = d/t
∴ t = d/v
Substituting in the above values in the given equation
t = 88/28
= 3.142 s
The acceleration is given by the formula
a = (v-u)/t
= (0 - 28)/3.142
= -8.912 m/s²
The negative sign is that the car is decelerating.
Hence, acceleration a = -8.912 m/s²
Answer:
1.791 MN
Explanation:
Thrust of the rocket can be found using the relation
T = v.dm/dt, where
T = thrust off the rocket
v = speed of the rocket, 9 km/s = 9000 m/s
dm/dt = rate at which fuel burns, 199 kg/s
Substituting the values into the formula, we have
T = 9000 * 199
T = 1791000 N
T = 1.791*10^6 N
Since 1 MN = 10^6, thus
T = 1.791 MN
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
No a wave with a high pitch has a very high frequency
