consider the motion of the tennis ball in downward direction
Y = vertical displacement = 400 m
a = acceleration = acceleration due to gravity = 9.8 m/s²
v₀ = initial velocity of the ball at the top of building = 10 m/s
v = final velocity of the ball when it hits the ground = ?
using the kinematics equation
v² = v²₀ + 2 a Y
inserting the values
v² = 10² + 2 (9.8) (400)
v = 89.11 m/s
Answer:
Yes
Explanation:
It is possible for sedimentary rocks to be converted to igneous rocks. Under conditions of high temperature and pressure, sedimentary rocks can be broken down into igneous rock by melting this rock type.
When the rock is broken down, it forms melt which when cooled and solidifies will form igneous rocks.
Sedimentary rocks are formed from the breaking down of pre-existing rocks through the action of weathering, erosion and sediment transportation. Within a basin, the sediments are compacted and lithified.
When this is subjected to intense pressure and temperature, the rock hardens and might further break down to melt.
Answer:
First one, third one, and fourth one
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

So the maximum acceleration on the block is

4)As the box slides, it is now subjected to kinetic friction, which is

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2
Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil,
= 926 kg/m³
density of the wood,
= 974 kg/m³
density of water,
= 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
![F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm](https://tex.z-dn.net/?f=F_%7Bwood%7D%20-%20F_%7Boil%7D%20-%20F_%7Bwater%7D%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.gh%20-%20%5Crho%20_o%20.g%28h-x%29%20-%20%5Crho_w%20.gx%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%20%5Crho%20_o%20%28h-x%29%20-%20%5Crho_w%20.x%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%5Crho%20_o%20h%20%2B%20%5Crho%20_o%20x%20-%20%5Crho_w%20.x%20%3D0%5C%5C%5C%5Ch%20%28%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%29%20%3D%20x%28%20%5Crho_w%20-%20%5Crho%20_o%29%5C%5C%5C%5Cx%20%3Dh%5B%5Cfrac%7B%20%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%7D%7B%5Crho_w%20-%20%5Crho%20_o%7D%20%5D%5C%5C%5C%5Cx%20%3D%203.69%5C%20cm%20%5Ctimes%20%5B%5Cfrac%7B974%20-%20926%7D%7B1000-926%7D%20%5D%5C%5C%5C%5Cx%20%3D%202.39%20%5C%20cm)
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.