Answer:
The approximate displacement of the object is <u>23 </u>m.
Explanation:
Given that:
v = 4t + 5 (m/s) for 3< t< 7; n= 4
The approximate displacement of the object can be calculated as follows:
The velocities at the intervals of t are :
3
4
5
6
the velocity at the intervals of t = 7 will be left out due the fact that we are calculating the left endpoint Reimann sum
n = 4 since there are 4 values for t, Then there is no need to divide the velocity values
v(3) = 4(3)+5
v(3) = 12+5
v(3) = 17
v(4)= 4(4)+5
v(4) = 16 + 5
v(4) = 21
v(5)= 4(5)+5
v(5) = 20 + 5
v(5) = 25
v(6) = 4(6)+5
v(6) = 24 + 5
v(6) = 29
Using Left end point;
= 23 m
Answer:
Part c)
Explanation:
Part a)
As per Coulomb's law we know that force on a charge placed in electrostatic field is given as
now acceleration of charge is given as
now if charge moved through the distance D in electric field and its speed changes from vo to fraction f of speed of light c
then we will have
so we have
Part b)
Now using work energy theorem we can say that total work done by electric force on moving charge will convert into kinetic energy
So we will have
so we have
Part c)
Now if an electron is accelerated using this field
then we have
so we have
Answer:
The distance the log has moved by the time Ernie reaches Bur is 1.33 m.
Explanation:
give information:
The log is 3.0 m long and has mass 20.0 kg.
Burt has mass 30.0 kg; Ernie has mass 40.0 kg
Ernie has mass 40.0 kg.
to find the distance, first, we have to calculate the center of mass
X = ∑ m x /∑m
= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)
= 150/90
= 5/3
when Ernie walk, the center of the mass is
X = (70 x 0) + (20 x (3/2))/(70 + 20)
= 30/90
= 1/3
the distance of log moved = 5/3 - 4/3 = 1.33 m
Answer:
The magnitude of the vertical component of its velocity as it hits the ground is 30 m/s. The horizontal distance from the firing point is 750 meters.
Air resistance is a type of friction.