Answer:
Between 120 and 180 seconds
Inyunirnhihirthithjitejreigjerwig
Answer:
The net torque on the square plate is 2.72 N-m.
Explanation:
Given that,
Side = 0.2 m
Force ![F_{1}=18\ N](https://tex.z-dn.net/?f=F_%7B1%7D%3D18%5C%20N)
Force ![F_{2}=26\ N](https://tex.z-dn.net/?f=F_%7B2%7D%3D26%5C%20N)
Force ![F_{3}=14\ N](https://tex.z-dn.net/?f=F_%7B3%7D%3D14%5C%20N)
We need to calculate the torque due to force F₁
Using formula of torque
![\tau_{1}=-F_{1}d_{1}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-F_%7B1%7Dd_%7B1%7D)
![\tau_{1}=-F_{1}\times\dfrac{a}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-F_%7B1%7D%5Ctimes%5Cdfrac%7Ba%7D%7B2%7D)
Put the value into the formula
![\tau_{1}=-18\times\dfrac{0.2}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-18%5Ctimes%5Cdfrac%7B0.2%7D%7B2%7D)
![\tau_{1}=-1.8\ N-m](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-1.8%5C%20N-m)
We need to calculate the torque due to force F₂
Using formula of torque
![\tau_{2}=F_{2}d_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3DF_%7B2%7Dd_%7B2%7D)
![\tau_{2}=F_{2}\times\dfrac{a}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3DF_%7B2%7D%5Ctimes%5Cdfrac%7Ba%7D%7B2%7D)
Put the value into the formula
![\tau_{2}=26\times\dfrac{0.2}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3D26%5Ctimes%5Cdfrac%7B0.2%7D%7B2%7D)
![\tau_{2}=2.6\ N-m](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3D2.6%5C%20N-m)
We need to calculate the torque due to force F₃
Using formula of torque
![\tau_{3}=F_{3}d_{3}](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3DF_%7B3%7Dd_%7B3%7D)
![\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3D%28F_%7B3%7D%5Csin45%2BF_%7B3%7D%5Ccos45%29%5Ctimes%5Cdfrac%7Ba%7D%7B2%7D)
Put the value into the formula
![\tau_{3}=0.1(14\sin45+14\cos45)](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3D0.1%2814%5Csin45%2B14%5Ccos45%29)
![\tau_{3}=1.92\ N-m](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3D1.92%5C%20N-m)
We need to calculate the net torque on the square plate
![\tau=\tau_{1}+\tau_{2}+\tau_{3}](https://tex.z-dn.net/?f=%5Ctau%3D%5Ctau_%7B1%7D%2B%5Ctau_%7B2%7D%2B%5Ctau_%7B3%7D)
![\tau=-1.8+2.6+1.92](https://tex.z-dn.net/?f=%5Ctau%3D-1.8%2B2.6%2B1.92)
![\tau=2.72\ N-m](https://tex.z-dn.net/?f=%5Ctau%3D2.72%5C%20N-m)
Hence, The net torque on the square plate is 2.72 N-m.
Answer:
![9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=9.4%5Ccdot%2010%5E%7B10%7D%20m)
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
![\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}](https://tex.z-dn.net/?f=%5Cfrac%7Br_a%5E3%7D%7BT_a%5E2%7D%3D%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7D)
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for
, we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.
weight of man = 75kg
speed of elevator, a = 1.5 ![m/ s^{2}](https://tex.z-dn.net/?f=m%2F%20s%5E%7B2%7D)
![F - w = ma \\](https://tex.z-dn.net/?f=F%20-%20w%20%3D%20ma%20%5C%5C)
![F = w + ma](https://tex.z-dn.net/?f=F%20%3D%20w%20%2B%20ma)
![F = m ( a +g )](https://tex.z-dn.net/?f=F%20%3D%20m%20%28%20a%20%2Bg%20%29)
![F = 75 ( 1.5 + 10 ) \\](https://tex.z-dn.net/?f=F%20%3D%2075%20%28%201.5%20%2B%2010%20%29%20%5C%5C)
![F = 75 ( 11.5 )](https://tex.z-dn.net/?f=F%20%3D%2075%20%28%2011.5%20%29)
![F = 862.5 N](https://tex.z-dn.net/?f=F%20%3D%20862.5%20N)
So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.
Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.
Learn more about speed here:-
brainly.com/question/19127881
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