Question 11
1 answer:
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The acceleration of the object is
Explanation:
We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:
where
F is the net force
m is the mass of the object
a is its acceleration
For the object in this problem,
F = 500 N is the applied force
m = 75 kg is the force
Solving the equation for a, we find the acceleration:
Learn more about Newton's second law:
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Answer:
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
Explanation:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters(m)
Equivalence
1nC= 10⁻⁹C
Data
K=8.99x10⁹N*m²/C²
q₁ = 7.94-nC= 7.94*10⁻⁹C
q₂= 4.14-nC= 4.14 *10⁻⁹C
d= 1.77 m
Magnitude of the electrostatic force that one charge exerts on the other
We apply formula (1):
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
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.
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
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.