Answer:
a= 3.49 m/s^2
Explanation:
magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.
we know that tangential acceleration a_t= change in velocity /time taken
now 90 km/h = 25 m/s
a_t = 25/17 = 1.47 m/s^2.
radial acceleration a_r = v^2/r
v= a_t×t = 1.47×13 = 19.11 m/s
a_r = 19.11^2/115= 3.175
now,
a= 3.49 m/s^2
Answer:
Q_2 = +/- 295.75*Q
Explanation:
Given:
- The charge of the first particle Q_1 = +Q
- The second charge = Q_2
- The position of first charge x_1 = 2a
- The position of the second charge x_2 = 13a
- The net Electric Field produced at origin is E_net = 2kQ / a^2
Find:
Explain how many values are possible for the unknown charge and find the possible values.
Solution:
- The Electric Field due to a charge is given by:
E = k*Q / r^2
Where, k: Coulomb's Constant
Q: The charge of particle
r: The distance from source
- The Electric Field due to charge 1:
E_1 = k*Q_1 / r^2
E_1 = k*Q / (2*a)^2
E_1 = k*Q / 4*a^2
- The Electric Field due to charge 2:
E_2 = k*Q_2 / r^2
E_2 = k*Q_2 / (13*a)^2
E_2 = +/- k*Q_2 / 169*a^2
- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:
E_net = E_1 + E_2
2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2
- The two equations are as follows:
1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2
2Q = Q / 4 + Q_2 / 169
Q_2 = 295.75*Q
2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2
2Q = Q / 4 - Q_2 / 169
Q_2 = -295.75*Q
- The two possible values corresponds to positive and negative charge Q_2.
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>
Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)
<h3>Resultant speed of the ball and crosswind</h3>
<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671
A non-example of force would be something that stay sill like a balloon in the air..... taste, smell, feel, texture, color, opinion, faith, hope, sincerity, honest, speed, momentum, altitude, volume, loudness, area, length, acidity, obesity, nationalism, current, resistance, viscosity, wavelength, flow, rate, frequency, albedo, diameter, age, temperature, acceleration, body mass index, salinity, specific, specific gravity, consciousness, intelligence, refraction index, mass, time, date rate, switching speed, libido, focal length, and latency are not force. And even there are many other things that also are not force, too.
Answer:
a)-1.014x 
J
b)3.296 x J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x 
x 47 x 110/ 3.4 => -1.014x J
b) at d= 0.8m (3.4-2.6) and =-1.014x J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+ = 
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore is zero
= +
The final potential energy is
= - GMaMb/d
Solving for ' '
= + GMaMb/d => -1.014x + 6.67 x x 47 x 110/ 0.8
= 3.296 x J
