Explanation :
It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
We know that acceleration is defined as the rate of change of velocity.
Where
dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s
dt is the change in time, dt = 40 s - 30 s = 10 s
So, 
From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. 
.
Answer:
u=36.8m/s
Explanation:
because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations
u^2=v^2-2ā*s. where:
u^2 stands for intial velocity
v^2 stands for final velocity
since the cougar skidded to a complete stop the final velocity is zero.
u^2=v^2-2ā*s
u^2=(0)^2 -2(-2.87 m/s^2)*236 m
u^2=0+5.74m/s^2* 236m
u^2=1354.64m^2/s^2
u=√1354.64m^2/s^2
u=36.8m/s (approximate value)
when ever the acceleration is constant you can use one of the following equation to find the required value.
1. v = u + at. (no s)
2. s= 1/2(u+v)t. (no ā)
3. s=ut + 1/2at^2. ( no v)
4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)
Answer:
Cools ; size
Explanation:
The rate at which magma cools determines the size of the crystals in the new rock. Igneous rocks are formed from the cooling and solidification of molten magma which finds its way to the surface or depth of very low pressure beneath the surface. This place or depth of cooling of magma affects the cooling rate and hence the size of the crystals formed. Igneous rocks formed at depths below the surface have more time to cool and allows more time for Crystal growth and hence produce coarse grained crystal grains called Intrusive igneous rocks which have significantly larger crystals than those formed on the surface which cools rapidly and allowing very little time for crystal growth giving rise to the formation of fine grained crystals and are called extrusive igneous rocks.
Answer:
The current through the inductor at the end of 2.60s is 9.7 mA.
Explanation:
Given;
emf of the inductor, V = 41.0 mV
inductance of the inductor, L = 13 H
initial current in the inductor, I₀ = 1.5 mA
change in time, Δt = 2.6 s
The emf of the inductor is given by;
Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.
I believe the answer is a
