Answer: A projectile is any object in which the only force is gravity
Explanation: Equations on how to calculate projectile velocity is stated below:
The initial velocity Vo being a vector quantity, has two componentsVox and Voy
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The acceleration A is a also a vector with two components Axand Ay given
Ax = 0 and Ay = - g = - 9.8 m/s2
Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant
Vx = Vocos(θ)
Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g
Vy = Vo sin(θ) - g t
Along the x axis the velocity Vx is constant and therefore the component x of the displacement is
x = Vocos(θ) t
Along the y axis, the motion is of uniform acceleration and the y component of the displacement is
y = Vo sin(θ) t - (1/2) g t2
Answer:
Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun.
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Answer:
High speed optical communication technology
To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.
Explanation:
Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.
The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.
This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J