The observation point on Earth and the two stars form a triangle. The two sides of the triangle are 23.3 ly and 34.76 ly and their included angle is 76.04°. We can use the cos rule to find the third side, which is the distance between the two stars.
c² = a² + b² - 2abCos(C)
c² = (23.3)² + (34.76)² - 2(23.3)(34.76)Cos(76.04)
c = 36.88 light years.
Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.
Let h = the maximum height reached.
Let u = the vertical launch velocity.
Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s
The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m
Answer: 122.5 m
Answer:
v_2 = 2*v
Explanation:
Given:
- Mass of both charges = m
- Charge 1 = Q_1
- Speed of particle 1 = v
- Charge 2 = 4*Q_1
- Potential difference p.d = 10 V
Find:
What speed does particle #2 attain?
Solution:
- The force on a charged particle in an electric field is given by:
F = Q*V / r
Where, r is the distance from one end to another.
- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:
F_net = m*a
- Equate the two expressions:
a = Q*V / m*r
- The speed of the particle in an electric field is given by third kinetic equation of motion.
v_f^2 - v_i^2 = 2*a*r
Where, v_f is the final velocity,
v_i is the initial velocity = 0
v_f^2 - 0 = 2*a*r
Substitute the expression for acceleration in equation of motion:
v_f^2 = 2*(Q*V / m*r)*r
v_f^2 = 2*Q*V / m
v_f = sqrt (2*Q*V / m)
- The velocity of first particle is v:
v = sqrt (20*Q / m)
- The velocity of second particle Q = 4Q
v_2 = sqrt (20*4*Q / m)
v_2 = 2*sqrt (20*Q / m)
v_2 = 2*v
