This question involves the concept of the scalar product.
The magnitude of the scalar product will be "0".
<h3>SCALAR PRODUCT</h3>
The scalar product, also known as the dot product of the two vectors is given by the following formula:
where,
- A.B = Scalar product = ?
- |A| = Mangnitude of vector A = 1 unit
- |B| = Magnitude of Vector B = 1 unit
- θ = Angle between vectors = 90°
Therefore,
A.B = 0
Learn more about scalar product here:
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Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
Answer: 7.5 rev/s
Explanation:
We are given the angular velocity 
a helicopter's main rotor blades:
However, we are asked to express this in the International Systrm (SI) units. In this sense, the SI unit for time is second (
):
An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
A) The answer is 11.53 m/s
The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi
Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²
Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h
So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h
We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m
1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s
b) The answer is 6.78 m
The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE
Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²
Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h
KE = PE
1/2 m * v² = m * a * h
Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?
1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
