Answer:
2km
Explanation:
Given data
We are told that the direction traveled are
North>>>East>>>South
Hence the displacement is defined as the distance away from the initial position is
Initial position =18km
FInal position = 16km
The displacement = 18-16= 2km
Hence the displacement is 2km
Hi my friend, since momentum is always conserved without external forces, the momentum after the collosion will still be 0.06 kg*m/s. Hope it helps☺
Answer:
Explanation:
The total travel time Suzette experiences is the sum of the times in each hallway. Using
time = distance/speed
we can add the times.
(35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)
= 10 s + 40 s + 12 s
= 62 s
It takes Suzette 62 seconds to get to class. She does not beat the bell.
Answer:
DU = 375 Joules
Explanation:
Given the following data;
Quantity of heat = 500 Joules
Work done = 125 Joules
To find the change in internal energy;
Mathematically, the change in internal energy of a system is given by the formula;
DU = Q - W
Where;
- DU is the change in internal energy.
- Q is the quantity of energy.
- W is the work done.
Substituting into the formula, we have;
DU = 500 - 125
<em>DU = 375 Joules</em>
<span>22.5 newtons.
First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as:
E = 0.5mv^2
where
E = Energy
m = mass
v = velocity
Substituting known values and solving gives:
E = 0.5 3.06 kg (7 m/s)^2
E = 1.53 kg 49 m^2/s^2
E = 74.97 kg*m^2/s^2
Now ignoring air resistance, how much energy should the rock have had?
We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So
3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2
So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is
299.88 J - 74.97 J = 224.91 J
So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division.
224.91 J / 10 m
= 224.91 kg*m^2/s^2 / 10 m
= 22.491 kg*m/s^2
= 22.491 N
Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
