Magnitude of the force of tension: 139 N
Explanation:
The surface of the ramp here is assumed to be the positive x-direction.
To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.
There are three forces acting along the x-direction:
- The force of tension,
, acting up along the plane - The force of friction,
, acting down along the plane - The component of the weight in the x-direction,
, acting down along the plane
We know that the magnitude of the weight is
So its x-component is
The net force along the x-direction can be written as
And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:
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Answer:
Explanation:
Generally, length of vector means the magnitude of the vector.
So, given a vector
R = a•i + b•j + c•k
Then, it magnitude can be caused using
|R|= √(a²+b²+c²)
So, applying this to each of the vector given.
(a) 2i + 4j + 3k
The length is
L = √(2²+4²+3²)
L = √(4+16+9)
L = √29
L = 5.385 unit
(b) 5i − 2j + k
Note that k means 1k
The length is
L = √(5²+(-2)²+1²)
Note that, -×- = +
L = √(25+4+1)
L = √30
L = 5.477 unit
(c) 2i − k
Note that, since there is no component j implies that j component is 0
L = 2i + 0j - 1k
The length is
L = √(2²+0²+(-1)²)
L = √(4+0+1)
L = √5
L = 2.236 unit
(d) 5i
Same as above no is j-component and k-component
L = 5i + 0j + 0k
The length is
L = √(5²+0²+0²)
L = √(25+0+0)
L = √25
L = 5 unit
(e) 3i − 2j − k
The length is
L = √(3²+(-2)²+(-1)²)
L = √(9+4+1)
L = √14
L = 3.742 unit
(f) i + j + k
The length is
L = √(1²+1²+1²)
L = √(1+1+1)
L = √3
L = 1.7321 unit
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
Answer:
True
Explanation:
Convection is a form of heat transfer that is predominantly common in fluids especially liquid and gas.
It occurs by the movement of a part of substance from one place to another based on density and temperature differences.
A typical convection cell is made up of a liquid that is heated. The liquid part close to the heat source becomes warmer and rises due to its low density. The part away from the heat source is more dense and begins to sink.
This analogy is commonly demonstrated in a boiling pot of water.
