Answer:
The the maximum force acting on the crate is 533.12 newtons.
Explanation:
It is given that,
Mass of the wooden crate, m = 136 kg
The coefficient of static friction, ![\mu_s=0.4](https://tex.z-dn.net/?f=%5Cmu_s%3D0.4)
The coefficient of kinetic friction, ![\mu_k=0.2](https://tex.z-dn.net/?f=%5Cmu_k%3D0.2)
We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :
![F=\mu_s mg](https://tex.z-dn.net/?f=F%3D%5Cmu_s%20mg)
![F=0.4\times 136\ kg\times 9.8\ m/s^2](https://tex.z-dn.net/?f=F%3D0.4%5Ctimes%20136%5C%20kg%5Ctimes%209.8%5C%20m%2Fs%5E2)
F = 533.12 N
So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.
Answer:
Its period if its length is increased by a factor of four is 5 s.
Explanation:
The period of a simple pendulum is given by;
![T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g} \ be \ constant \\\\\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D%20%5C%5C%5C%5C%5Cfrac%7BT%7D%7B2%5Cpi%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D%20%5C%5C%5C%5C%20%5Cfrac%7BT%5E2%7D%7B4%5Cpi%5E2%7D%20%3D%20%5Cfrac%7Bl%7D%7Bg%7D%5C%5C%5C%5C%5Cfrac%7BT%5E2%7D%7Bl%7D%20%3D%20%5Cfrac%7B4%5Cpi%5E2%7D%7Bg%7D%20%5C%5C%5C%5Clet%20%5C%20%5Cfrac%7B4%5Cpi%5E2%7D%7Bg%7D%20%20%5C%20be%20%5C%20constant%20%5C%5C%5C%5C%5Cfrac%7BT_1%5E2%7D%7Bl_1%7D%20%20%3D%20%5Cfrac%7BT_2%5E2%7D%7Bl_2%7D%20%5C%5C%5C%5C)
Given;
initial period, T₁ = 2.5
initial length, = L₁
new length, L₂ = 4L₁
the new period, T₂ = ?
![\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\ T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\ T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%5E2%7D%7Bl_1%7D%20%20%3D%20%5Cfrac%7BT_2%5E2%7D%7Bl_2%7D%20%5C%5C%5C%5CT_2%5E2%20%3D%20%5Cfrac%7BT_1%5E2%20l_2%7D%7Bl_1%7D%20%5C%5C%5C%5CT_2%20%3D%20%5Csqrt%7B%5Cfrac%7BT_1%5E2%20l_2%7D%7Bl_1%7D%7D%20%5C%5C%5C%5C%20%20T_2%20%3D%20%5Csqrt%7B%5Cfrac%7B%282.5%29%5E2%20%5C%20%5Ctimes%20%5C%204l_1%7D%7Bl_1%7D%7D%5C%5C%5C%5C%20%20T_2%20%3D%5Csqrt%7B%282.5%29%5E2%20%5C%20%5Ctimes%20%5C%204%7D%5C%5C%5C%5CT_2%20%3D%20%5Csqrt%7B25%7D%20%5C%5C%5C%5CT_2%20%3D%205%5C%20s)
Therefore, its period if its length is increased by a factor of four is 5 s.
Answer:
YES
Explanation:
An echo may be defined as a sound which is repeated because of the sound waves that are produced are reflected back after striking a surface. Sound waves can smoothly bounce off the hard objects in the same manner as a rubber ball bounces back the ground.
When a sound wave strikes a hard surface, the sound waves gets reflected back and bounces back to the observer and produces an echo. If the sound waves strikes a soft surface it absorbs the sound.
Although the direction of a sound changes but the echo sounds in the same way as the original sound.
D because u r using your energy to do all the things
Answer:
Vectors and arrows.
Explanation:
Mathematically speaking, forces are represented by vectors, which have magnitude and direction. Graphically speaking, vectors are represented by arrows.