The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.
The given parameters;
- magnitude of force on the object, F = 10 N
- angle between the object and the horizontal flat surface = 0⁰
Apply Newton's second law of motion to determine the magnitude of the force on the object.
Due to the position of the object, the magnitude of the force acting on it is calculated as;
Therefore, the magnitude of the force acting on the object is 10 N.
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Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector
.
The positive y-axis = the northern direction, with unit vector
.
The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
![\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B2%7D%20%3D40%28cos%2834%5E%7Bo%7D%29%5Chat%7Bi%7D%20-%20sin%2824%5E%7Bo%7D%29%5D%5Chat%7Bj%7D%29%20%3D%2033.1615%5Chat%7Bi%7D%20-22.3677%5Chat%7Bj%7D%29)
The plane's actual velocity is the vector sum of the two velocities. It is
The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h
The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°
Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.
With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.
Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2
Then the equation for velocity, V is
V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t
Vo is the initial velocity, which you can find using V = 54km/h and t = -2
Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h
Then, the equation is: V = 50 km/h + 2t
Valid for constant acceleration.
Answer:
0.8
Explanation:
The two spheres have the same potential, V.
Let the radius of the larger sphere be R and the radius of the smaller sphere be r,
=> R = 4r
Let the charge on the smaller sphere be q. Hence, the larger sphere will have charge Q - q.
The potential of the smaller sphere will be:
The potential of the larger sphere will be:
Inputting R = 4r,
Since 
,
=> Q - q = 4q
=> 5q = Q
q = 0.2Q
The fraction of the charge Q that rests on the smaller sphere is 0.2
The charge of the larger sphere is:
Q - q = Q - 0.2Q = 0.8Q
∴ The fraction of the total charge Q that rests on the larger sphere is 0.8
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
