Answer:
<u>B</u><u>.</u><u> </u><u>Transverse</u><u> </u><u>wave</u><u>.</u>
Explanation:
Because it has troughs and crests.
Answer:
the mass should be bring closer to the point about which we are finding torque
Explanation:
τ = Σr × F = rmg
where m is the mass, g is acceleration due to gravity, and r is the distance
Torque is directly proportional to -
1.mass, m , of object
2. distance, r, of the mass from the point about which we are finding the torque.
So if we increase or decrease them then the torque will also increase or decrease.
So if we increase the mass the torque will increase but since we have to maintain same torque therefore we have to decrease the distance of mass from the point about which we are finding torque.
Therefore the mass should be bring closer to the point about which we are finding torque.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
Answer:
L = 41.09 Kg m2 / s The angular momentum does not depend on the time
Explanation:
The definition of angular momentum is
L = r x p
Where blacks indicate vectors
Let's apply this definition our case. Linear momentum
p = m v
Let's replace
L = m r x v
The given function is
x = 6.00 i ^ + 4.15 t j
^
We look for speed
v = dx / dt
v = 0 + 4.15 j ^
To evaluate the angular momentum one of the best ways is to use determinants
![L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]](https://tex.z-dn.net/?f=L%20%3D%20m%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C6%264.15t%260%5C%5C0%264.15%260%5Cend%7Barray%7D%5Cright%5D)
L = m 6 4.15 k ^
The other products give zero
Let's calculate
L = 1.65 6 4.15 k ^
L = 41.09 Kg m2 / s
The angular momentum does not depend on the time
Answer:
The net power needed to change the speed of the vehicle is 275,000 W
Explanation:
Given;
mass of the sport vehicle, m = 1600 kg
initial velocity of the vehicle, u = 15 m/s
final velocity of the vehicle, v = 40 m/s
time of motion, t = 4 s
The force needed to change the speed of the sport vehicle;
The net power needed to change the speed of the vehicle is calculated as;
![P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W](https://tex.z-dn.net/?f=P_%7Bnet%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20F%5Bu%20%2B%20v%5D%5C%5C%5C%5CP_%7Bnet%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%2C000%5B15%20%2B%2040%5D%5C%5C%5C%5CP_%7Bnet%7D%20%3D%20275%2C000%20%5C%20W)
