Work=force*displacement=(588.6 N)(1.2 m)=706.32 which is about 700 J for sig figs
displacement=1.2 m
force (needs to equal gravity so it can counteract gravity)=m g=(60kg)(9.81 m/s^2)=588.6 N
Some very wise man noticed that the shadow cast from a stick placed vertically in the ground was a different angle at exactly noon than the angle of shadow cast at exactly noon from a vertically placed stick in the ground at a location further north. By employing mathematics, he was able to deduce that the earth is in fact, roundish.
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant before it hits the ground?
After 3.5 seconds of free-fall on or near the surface of the Earth, (ignoring effectsof air resistance), the vertical speed of an object starting from rest isg T = 3.5 g = 3.5 x 9.8 = 34.3 meters per second.With no initial horizontal component, the direction of such an object's velocitywhen it hits the ground is straight down.
The question has no answer.This sort of question is based on the assumption that gravitational acceleration is 9.8 m/s^2 (downwards). In that case, if the rock is projected from the ground its height at time t is 24t - 4.9t^2. It cannot be greater than that unless air resistance is negative!
if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...
An object thrown vertically up wards from the ground returned back to the ground in 6s after it was thown up if it reached a height of 12m calculate?
To determine how far the car fell vertically from the bridge to the ground you would need to know how far the bridge roadway is from the ground.
Water moves in three directions: downward due to gravity, horizontally as surface runoff, and vertically through infiltration into the ground.
The velocity of the penny as it hits the ground can be calculated using the equation: velocity = distance/time. Assuming the penny falls vertically, if we take the distance it falls to be 9.8 m/s^2 x (4.5 s)^2 / 2 ≈ 99.22 meters and the time is 4.5 seconds, the velocity would be 99.22 meters / 4.5 seconds = 22.04 m/s.
Because a box has 6 faces, the "front surface" is not detailed enough. If you are referring to the surface perpendicular to the ground and is facing you, then you simply multiply the height of the box by the edge of the base on the front surface.
Surface waves are caused by the interaction between seismic waves and the Earth's surface. These waves travel along the Earth's crust and can cause the ground to shake horizontally and vertically. Surface waves are typically the most destructive type of seismic waves during an earthquake.
A ground-hugging stem is a type of stem that grows close to the ground or trails along the surface. It allows plants to spread out horizontally rather than grow vertically, helping them to creep or spread across the ground. These types of stems are often found in plants like strawberries or certain types of vines.
without the rest of that problem, we can't even give you a formula to calculate your answer
To calculate the velocity of the ball, we need to know the height from which it was dropped. If the ball was dropped from rest, we can use the formula for free fall motion: velocity = (acceleration due to gravity * time). Assuming the acceleration due to gravity is 9.81 m/s^2, the velocity of the ball hitting the ground after 3.03 seconds would be around 29.7 m/s.
In physics, slope refers to the steepness of a surface, usually measured as the angle between the surface and the horizontal ground. It is used to calculate the force of gravity acting on an object on an inclined surface and to determine the acceleration and motion of objects on slopes.
The answer depends on whether the ball is thrown vertically upwards or downwards. That critical piece of information is not provided!
Yes!! Since the ground is a solid surface, it is relatively easy to anchor the "zero" of the measuring tape to that surface. Then the tape can be extended vertically directly next to the net, and the height of the net can be viewed (standing on a ladder) without touching the net. If you tryo to anchor the "zero" to the top of the net and measure down to the ground, there is a very good chance that the top of the net will be pulled downward, and its actual height will be inaccurate.