Answer:
2. Using a shorter string of length L ′ ≈ 0.25 meters
5. Using a shorter string of length L ′ ≈ 0.5 meters
Explanation:
The period of a pendulum is given by
where
L is the length of the pendulum
g is the acceleration due to gravity
We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.
Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.
Therefore, the only alterations that will reduce the period of the pendulum are:
2. Using a shorter string of length L ′ ≈ 0.25 meters
5. Using a shorter string of length L ′ ≈ 0.5 meters
Not so fast.
I think you're using 'accelerating' to mean 'speeding up', but you really need
to be more careful with it. "Acceleration" means ANY change in speed OR
direction.
If an object's speed to the left is decreasing, or its speed to the right is
increasing, then the net force on the object must be directed towards
the right.
If an object is moving with constant speed in a circular path, then it's
constantly accelerating, because its direction is constantly changing.
The force on it is always directed towards the center of the circle, so
there's one point on the path where the force is directed straight to the right.
Answer: 1 How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-. 2 34 °C? 4. How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of ... A saturated solution of KClO3 was made with 300 g of H2O at. 34 °C.
Explanation:
Answer:
Explanation:
Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F
a ) Frictional force = μ R = F cosφ
R = mg - F sinφ
μ(mg - F sinφ) = F cosφ
μmg = F (μsinφ+cosφ)
F = μmg / (μsinφ+cosφ)
Work done
= F cosφ x d
= μmg x cosφ x d / (μsinφ+cosφ)
b )Work done
= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)
= 1164.61 / .87946
1324.23 J
c ) work done on the sled by friction
= - (work done by force)
= - μmg x cosφ x d / (μsinφ+cosφ)
d ) work done on the sled by friction
= - 1324.23 J
Explanation:
what is asked give brief explanation about question
