In BPC
tan\theta =a/b = 3/4
\theta = tan^-1(0.75)
\theta = 36.87 deg
BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m
Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C
Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C
Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C
Net electric field along X-direction is given as
Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C
Net electric field along X-direction is given as
Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C
Net electric field is given as
E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
Answer:
A) 350 N
B) 58.33 N
C) 35 kg
D) 35 kg
Explanation:
If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2
The weight of the object on Earth is given by:
Weight = mass * g = 35 * 10 = 350 N
The weight of the object on the Moon:
Weight = mass * gmoon = 35 * 5/3 = 58.33 N
The mass of the object on Earth is 35 kg
The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.
Answer:
The height of the bridge is 78.4 m.
Explanation:
Given;
time of the stone motion off the bridge, t = 4.0 s
acceleration due to gravity, g = 9.8 m/s²
The height of the bridge is given by;
h = ut + ¹/₂gt²
where;
u is the initial velocity of the stone, u = 0
h = ¹/₂gt²
h = ¹/₂(9.8)(4)²
h = 78.4 m
Therefore, the height of the bridge is 78.4 m.
Answer:
Explanation:
Since the system is in international space station
so here we can say that net force on the system is zero here
so Force by the astronaut on the space station = Force due to space station on boy
so here we know that
mass of boy = 70 kg
acceleration of boy = 
now we know that
now for the space station will be same as above force
Answer:
14m/s
Explanation:
Given parameters:
Radius of the curve = 50m
Centripetal acceleration = 3.92m/s²
Unknown:
Speed needed to keep the car on the curve = ?
Solution:
The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.
It is given as;
a = 
a is the centripetal acceleration
v is the speed
r is the radius
Now insert the parameters and find v;
v² = ar
v² = 3.92 x 50 = 196
v = √196 = 14m/s
