Answer:
The magnitude of the rate of change of the child's momentum is 794.11 N.
Explanation:
Given that,
Mass of child = 27 kg
Speed of child in horizontal = 10 m/s
Length = 3.40 m
There is a rate of change of the perpendicular component of momentum.
Centripetal force acts always towards the center.
We need to calculate the magnitude of the rate of change of the child's momentum
Using formula of momentum
Put the value into the formula
Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.
Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.
Answer:
The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.
Explanation:
This is a horizontal bar.
For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.
Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:
Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:
So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).
Answer:
(a)F= 3.83 * 10^3 N
(b)Altitude=8.20 * 10^5 m
Explanation:
On the launchpad weight = gravitational force between earth and satellite.
W = GMm/R²
where R is the earth radius.
Re-arranging:
WR² / GM = m
m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg
The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:
Fc = mω²r
where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.
ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s
When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force between the earth and the satellite:
Fc = GMm/r²
mω²r = GMm / r²
ω²r = GM / r²
r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²
r³ = 3.612 * 10^20
r = 7.12 * 10^6 m
(a)
F = GMm/r²
F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²
F= 3.83 * 10^3 N
(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m
Answer:
1.) 6 hectokisses 50 times
2.) Millionaire
Explanation:
