Answer:
a. -369.36J
b. -123.9J
c. 9.52m
Explanation:
From the expression for kinetic energy
K. E=1/2mv^2
Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is
K.E=1/2*19(3.6²-7.2²)
K.E= -369.36J
b. to determine the workdone by the force,we determine the distance moved.
But the acceleration is from
F=ma ,
a=f/m
a=-13/19
0.68m/s²
the distance moved is
s=v²/2a
s=3.6²/2*0.68
s=9.52m
Hence the work done is
W=force * distance
W=-13*9.52
W=-123.9J
d. the distance moved is
s=v²/2a
s=3.6²/2*0.68
s=9.52m
k = 5.29
a = 0.78m/s²
KE = 0.0765J
<u>Explanation:</u>
Given-
Mass of air tracker, m = 1.15kg
Force, F = 0.9N
distance, x = 0.17m
(a) Effective spring constant, k = ?
Force = kx
0.9 = k X0.17
k = 5.29
(b) Maximum acceleration, m = ?
We know,
Force = ma
0.9N = 1.15 X a
a = 0.78 m/s²
c) kinetic energy, KE of the glider at x = 0.00 m.
The work done as the glider was moved = Average force * distance
This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0
As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)
Work = Kinetic energy
KE = 0.450 * 0.17
KE = 0.0765J
<span>d) While taking a shower, I saw a little mouse scurry across the bathroom floor.
This sentence is ordered properly. </span>
Answer:
a) 
degrees
b) 
degrees
c) 
degrees
Explanation:
If we have a vector 
in a two dimensional space. The angle respect the x axis can be founded from the following expression:
And then the angle is given by:
Part a
Ax = 11 m and Ay = 11 m
For this case the angle would be:
degrees
Part b
Ax = 19 m and Ay = 11 m
For this case the angle would be:
degrees
Part c
Ax = 11 m and Ay = 19 m
For this case the angle would be:
degrees
Answer:
Explanation:
We shall write the velocities given in vector form to make the solution easy.
The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form
waV(e) = 2.2 cos 60 i + 2.2 sin 60 j
waV(e) = 1.1 i + 1.9 j
Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows
wiV(e) = - 4.5 cos 50 i - 4.5 sin 50 j
wiV(e) = - 2.89 i - 3.45 j
Now we have to calculate velocity of wind with respect to water that is
wiVwa
wiV( wa) = wiV ( e)+ eV(wa)
= wiV( e)- waV(e)
- 2.89 i - 3.45 j - 1.1 i - 1.9 j
= - 3.99 i - 5.35 j
Magnitude of this relative velocity
D² = 3.99² + 5.35²
d = 6.67 m /s
