Answer:
The net force acting on an object is proportional to the acceleration of that object with respect to an inertial frame of reference. The constant of proportionality in this, Newton's second law, is the classical mass of the object.
Explanation:
Double
Explanation:
Since the period T of a pendulum is given by
![T = 2\pi \sqrt{\dfrac{l}{g}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cdfrac%7Bl%7D%7Bg%7D%7D)
By increasing the length of the pendulum by 4, the period becomes
![T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T](https://tex.z-dn.net/?f=T%27%20%3D%202%5Cpi%20%5Csqrt%7B%5Cdfrac%7B4l%7D%7Bg%7D%7D%20%3D%202%5Cleft%282%5Cpi%20%5Csqrt%7B%5Cdfrac%7Bl%7D%7Bg%7D%7D%5Cright%29%20%3D%202T)
You can see that the period doubles when we increase the length by a factor of 4.
Answer:
θ = 33°
Explanation:
Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:
![T = \frac{2\ u\ Sin\theta}{g} \\\\Sin\theta = \frac{Tg}{2u}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5C%20u%5C%20Sin%5Ctheta%7D%7Bg%7D%20%5C%5C%5C%5CSin%5Ctheta%20%3D%20%5Cfrac%7BTg%7D%7B2u%7D)
where,
θ = launch angle = ?
T = Total time of flight = 0.6 s
g = acceleration due to gravity = 9.81 m/s²
u = launch speed = 5.4 m/s
Therefore,
![Sin\theta = \frac{(0.6\ s)(9.81\ m/s^2)}{(2)(5.4\ m/s)}\\\\\theta = Sin^{-1}(0.545)](https://tex.z-dn.net/?f=Sin%5Ctheta%20%3D%20%5Cfrac%7B%280.6%5C%20s%29%289.81%5C%20m%2Fs%5E2%29%7D%7B%282%29%285.4%5C%20m%2Fs%29%7D%5C%5C%5C%5C%5Ctheta%20%3D%20Sin%5E%7B-1%7D%280.545%29)
<u>θ = 33°</u>
Linear momentum has to be conserved. It was zero before the thread eas burned ... when nothing was moving ... so the momentum of the masses moving in opposite directions has to add up to zero. ... Momentum = mass times speed. ... In one direction, you have 5 kg times 1/5 m/s= 1 kg-m/s. ... We need 1 kg-m/s in the other direction. ... 7 kg times speed = 1 kg-m/s. ... Can you finish it from here ?
Answer: He must drive at 58.8mi/hr
{3 significant figure}
Explanation: The original speed of the car to complete the journey normally is = 550mile/10hr
= 55mi/hr
The normal time to cover 120 mile with that speed is
t = distance/speed
= 120/55
=2hr, 0.18*60=11mins
= 2hr11mins.
But the driver found out he was 30mins behind at this distance, so he spent
2hr 11min +30min= 2hr41mins
For him to meet up the original schedule time 10hrs, he has
(10-2.41)hrs = 7hr19mins to cover the remaining distance of
{550-120}=430mile.
First, 7hr19mins= {7*(19/60)}hr
= {439/60}hr
Now let's find the speed at which the driver must move for him to cover for his initial delay
Speed= distance/time
= 430/{439/60}
This is same as writing
Speed = 430/439/60
Using the law of reciprocal,
Speed = (430*60)/439
= 25800/439
= 58.7699mi/hr
But we were asked to leave our answer in 3 significant figure. Therefore,
Speed = 58.8mi/hr