Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
2271.16N/C upward
Explanation:
The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.
First we determine the downward weight using
Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be
To calculate the electric field,
Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.
Hence the magnitude of the electric force is 2271.16N/C and the direction is upward
Answer:
At a constant speed
Explanation:
If a car is going 30 mph and it isnt going faster or slower, it is not accelerating but it is still moving
Answer:
F = 352 N
Explanation:
we know that:
F*t = ΔP
so:
F*t = M
-M
where F is the force excerted by the wall, t is the time, M the mass of the ball, the final velocity of the ball and the initial velocity.
Replacing values, we get:
F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)
solving for F:
F = 352 N
Answer:
Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10
3
)×(9.5×60)=14.25×10
5
J
Energy usefully consumed =msΔT=3×(4.2×10
3
)×(100−15)=10.71×10
5
where s=4.2J/g
o
C= specific heat of water and boiling point temp=100
o
C
Heat lost =14.25×10
5
−10.71×10
5
=3.54×10
5
