Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
A psychologist who would claim that a client's personal experience and viewpoint influence behavior more than events in reality would probably use cognitive psychology mixed with developmental aspects to explain the behavior and personality of a person.
<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)
So, Your Answer would be Option B
Hope this helps!</span>
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
Answer:
68.8 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of box = 18 Kg
Coefficient of friction (μ) = 0.39
Force of friction (F) =?
Next, we shall determine the normal force of the box. This is illustrated below:
Mass (m) of object = 18 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal force (N) =?
N = mg
N = 18 × 9.8
N = 176.4 N
Finally, we shall determine the force of friction experienced by the object. This is illustrated below:
Coefficient of friction (μ) = 0.39
Normal force (N) = 176.4 N
Force of friction (F) =?
F = μN
F = 0.39 × 176.4
F = 68.796 ≈ 68.8 N
Thus, the box experience a frictional force of 68.8 N.
