v = a X t. a = 9.8 m/s2 (acceleration) and t = time.
At t = 2 seconds, v = (9.8 m/s2) X (2 s) = 19.8 m/s.
At t = 4 seconds, v = (9.8 m/s2) X (4 s) = 39.6 m/s.
[ANSWER: at t = 2 s, v = 19.8 m/s ; at t = 4 s, v = 39.6 m/s. ]
Escape velocity from Earth depends only on the mass of the Earth and the distance from its center, not the mass or size of the rocket. All rockets need to reach the same escape velocity to leave Earth's gravitational pull, regardless of their size.
No, an acorn is too small and light to reach its terminal velocity when falling from a tree. Terminal velocity is the maximum constant speed that an object reaches when the drag force equals the force of gravity, and the small size and weight of an acorn mean that it doesn't experience enough air resistance to reach this terminal velocity.
A small slope on a velocity vs. time graph indicates a small acceleration because acceleration is the rate of change of velocity. A small slope means the velocity is changing slowly, which corresponds to a small acceleration. The steeper the slope, the larger the acceleration.
Yes, a small slope on a velocity vs time graph indicates a small acceleration. The slope of a velocity-time graph represents the acceleration of an object. A small slope indicates a small change in velocity over time, which corresponds to a small acceleration.
A small slope on a velocity vs time graph indicates a small acceleration. This means that the velocity of the object is changing at a slow rate over time.
meteorite.
Escape velocity from Earth depends only on the mass of the Earth and the distance from its center, not the mass or size of the rocket. All rockets need to reach the same escape velocity to leave Earth's gravitational pull, regardless of their size.
No, an acorn is too small and light to reach its terminal velocity when falling from a tree. Terminal velocity is the maximum constant speed that an object reaches when the drag force equals the force of gravity, and the small size and weight of an acorn mean that it doesn't experience enough air resistance to reach this terminal velocity.
Generally it falls into the atmosphere and burns up.
Velocity, velocity, velocity. Both a falling apple and the orbiting Moon are being attracted by the Earth's gravity. The apple is still being attracted even when it is on the ground. The difference is that the Moon is moving laterally, that is, its motion has a tangential velocity component that carries it "past the edge of the Earth" as it falls, so that it never reaches the ground. It also has very little friction moving through space to slow it down. So gravity, in trying to pull it down, keeps it from flying off into space, but can never bring it crashing down on Isaac Newton's head. ========== Besides the tree is a wee bit small to hold the moon!
A small slope on a velocity vs. time graph indicates a small acceleration because acceleration is the rate of change of velocity. A small slope means the velocity is changing slowly, which corresponds to a small acceleration. The steeper the slope, the larger the acceleration.
Yes, a small slope on a velocity vs time graph indicates a small acceleration. The slope of a velocity-time graph represents the acceleration of an object. A small slope indicates a small change in velocity over time, which corresponds to a small acceleration.
A small slope on a velocity vs time graph indicates a small acceleration. This means that the velocity of the object is changing at a slow rate over time.
True
To find the recoil velocity of the Earth when a 5 kg bowling ball is projected upward with a velocity of 2.0 meters per second, we can use the principle of conservation of momentum. Initially, the total momentum is zero, so the momentum gained by the bowling ball must equal the momentum lost by the Earth. The momentum of the bowling ball is ( p = mv = 5 , \text{kg} \times 2 , \text{m/s} = 10 , \text{kg m/s} ). Since the mass of the Earth is approximately ( 5.97 \times 10^{24} , \text{kg} ), the recoil velocity of the Earth can be calculated as ( v_{Earth} = -\frac{p_{ball}}{m_{Earth}} = -\frac{10}{5.97 \times 10^{24}} \approx -1.67 \times 10^{-24} , \text{m/s} ), indicating an extremely small downward velocity.
That would depend on what you consider "large".The size of an object's momentum = (its mass) x (its speed).So, more mass and more speed result in more momentum.
It does. When you jump, both you and the Earth move apart, and then fall back together. For every action, there is an equal and opposite reaction. After the jump, the equation m(person)*v(person) = - m(earth)*v(earth). A typical person weights 70 kg, and the Earth weighs approximately 6,000,000,000,000,000,000,000,000 kg. So to make the equation equal, the velocity of the Earth is about 1/85000000000000000000000 as much as the velocity of the person. This is really too small to notice by any current means of measurement.