Answer:
2. mechanical weathering can produce smaller pieces of rock that have more surface area for chemical weathering to work Explanation:
Mechanical weathering involves activities of living organisms or some geological processes. The bigger rocks are usually reduced to smaller rocks and further reduction might be limited or not posibble mechanically. This reduced rocks now increases the surface area available for chemical weathering; which further reduces the sizes of the rocks below the size range of mechanical weathering. one will recall that the rate of chemical reaction increases with exposed surface area.
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
The appropriate response is Gallium. It is a concoction component with image Ga and nuclear number 31. It is in gathering 13 of the occasional table and subsequently has similitudes to alternate metals of the gathering, aluminum, indium, and thallium.
<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
