The solution for this problem is:
Remember that this doesn’t depend on the mass of the child.
E = T + U = constant
E (maximum height) = T + U =U = mgh = mg[r - r· cos (Θ)]
E (bottom height) = T + U = T = ½mv² = mg[r - r · cos (Θ)]
v² = 2g[r – r · cos (Θ)]
v = √ (2g[r-r·cos(Θ)])
= √(2(9.8)[3 – 3 · cos (45°)])
= 4.15 m/s or 15 kph
Answer:
a) 0.9995c
b) 5641MeV
c) 91670 MeV
Explanation:
(a) The speed of approach is given by the formula:
(b) the kinetic energy is given by:
![E_k=m_0c^2[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1]](https://tex.z-dn.net/?f=E_k%3Dm_0c%5E2%5B%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7D-1%5D)
by replacing c=3*10^8m/s, m_0=1.67*10^{-27}kg we obtain:
(c) in the rest frame of the other proton we have:
![E_k=m_0c^2[\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}-1]](https://tex.z-dn.net/?f=E_k%3Dm_0c%5E2%5B%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bu%5E2%7D%7Bc%5E2%7D%7D%7D-1%5D)
by replacing we get
hope this helps!!
Answer:
Work done, W = 141174 Joules
Explanation:
It is given that,
Constant tension acting on the boat, T = F = 465 N
Speed of the boat, v = 4.6 m/s
Time, t = 1.1 min = 66 seconds
Let W is the work done by the tension. It is equal to the product of force and displacement. It is given by :
Since, 
W = 141174 Joules
So, the work is done by the tension is 141174 Joules. Hence, this is the required solution.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
Answer:
by reducing the frequency of the wave by a factor of three
Explanation:
Speed = wavelength * frequency.
Wavelength = speed/frequency
The speed of an electromagnetic wave in a vacuum is constant. Meanwhile the wavelength and frequency have an inverse relationship which means for every rise in any of the parameters there will also be a corresponding fall.
Since the wavelength is tripled then the frequency will also be reduced by a factor of 3
