Answer:
a) the magnitude of r is 184.62
b) the direction is 37.74° south of the negative x-axis
Explanation:
Given the data in the question;
as illustrated in the image blow;
To find the the magnitude of r, we will use the Pythagoras theorem
r² = y² + x²
r = √( y² + x²)
we substitute
r = √((-113)² + (-146)²)
r = √(12769 + 21316 )
r = √(34085 )
r = 184.62
Therefore, the magnitude of r is 184.62
To find its direction, we need to find ∅
from SOH CAH TOA
tan = opposite / adjacent
tan∅ = -113 / -146
tan∅ = 0.77397
∅ = tan⁻¹( 0.77397 )
∅ = 37.74°
Therefore, the direction is 37.74° south of the negative x-axis
The question involves a ping-pong ball that is held submerged in a bucket by a string attached to the bottom of the bucket.
The answer is the tension of the string will increase. This is because making the water salty increases its density, and consequently, increases its buoyancy. This is why sea water is more buoyant than fresh water. Therefore the ping pong is pushed more upwards by the water when salt is added than initially. This gives the string more tension.
Answer:
Executive Orders state mandatory requirements for the Executive Branch, and have the effect of law. They are issued in relation to a law passed by Congress or based on powers granted to the President in the Constitution and must be consistent with those authorities.
Explanation:
Answer:
Explanation:
The combined wave only end up been more powerful than the Longitudinal wave. This means, the transverse wave is more powerful than the combined wave. In transverse wave, the oscillation is perpendicular to the direction of the wave, while in longitudinal wave, the motion of the movement of the object is parallel to the movement of the wave. And in combined wave, the movement of the medium is in a circular manner,
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃
We divide by (k * q3) on both sides of the equation
q₁ = + 1.25 nC
