Answer:
<h2>
206.67N</h2>
Explanation:
The sum of force along both components x and y is expressed as;
![\sum Fx = ma_x \ and \ \sum Fy = ma_y](https://tex.z-dn.net/?f=%5Csum%20Fx%20%3D%20ma_x%20%20%5C%20and%20%5C%20%5Csum%20Fy%20%3D%20ma_y)
The magnitude of the net force which is also known as the resultant will be expressed as ![R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}](https://tex.z-dn.net/?f=R%20%3D%5Csqrt%7B%28%5Csum%20Fx%29%5E2%20%2B%20%28%5Csum%20Fx%20%29%5E2%7D)
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
![a_x = \frac{d^2 x }{dt^2}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bd%5E2%20x%20%7D%7Bdt%5E2%7D)
![a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Cfrac%7Bdx%7D%7Bdt%7D%20%29%5C%5C%20%5C%5Ca_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%286t%5E%7B2%7D-4%20%20%29%5C%5C%5C%5Ca_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2812t%20%20%29%5C%5C%5C%5Ca_x%20%3D%2012m%2Fs%5E%7B2%7D)
Similarly,
![a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2](https://tex.z-dn.net/?f=a_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Cfrac%7Bdy%7D%7Bdt%7D%20%29%5C%5C%20%5C%5Ca_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%285t%5E%7B3%7D%20%2B6%20%29%5C%5C%5C%5Ca_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2815t%5E%7B2%7D%20%20%20%29%5C%5C%5C%5Ca_y%20%3D%2030t%5C%5Ca_y%20%5C%20at%20%5C%20t%3D%202.15s%3B%20a_y%20%3D%2030%282.15%29%5C%5Ca_y%20%3D%2064.5m%2Fs%5E2)
![\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N](https://tex.z-dn.net/?f=%5Csum%20F_x%20%3D%203.15%20%2A%2012%20%3D%2037.8N%5C%5C%5Csum%20F_y%20%3D%203.15%20%2A%2064.5%20%3D%20203.18N)
![R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B37.8%5E2%2B203.18%5E2%7D%5C%5C%20%5C%5CR%20%3D%20%5Csqrt%7B1428.84%2B41%2C282.11%7D%5C%5C%20%5C%5CR%20%3D%20%5Csqrt%7B42.710.95%7D%5C%5C%20%5C%5CR%20%3D%20206.67N)
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
15 miles to kilometers would be: 24.14 kilometers
Answer: Physical changes in nature could then be erosion in a mountain, the melting of snow, and a river freezing over from the cold. Since none of these changes affect the chemical composition of the mountain, the snow, or the river, they are physical changes.
Explanation:
Answer:
A
Explanation:
Resistors in series add. There is only one path the current can take. That's why Christmas Tree lights sometimes give a lot of trouble. If a bulb burns out, it could be any one of them and time is needed to find the burned out bulb.
That being the case R = R1 + R2
R1 = 50 ohms
R2 = 50 ohms
R = 50 + 50
R = 100 ohms
Answer A
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, ![v_t=60\ m/s](https://tex.z-dn.net/?f=v_t%3D60%5C%20m%2Fs)
Area of cross section, ![A=0.33\ m^2](https://tex.z-dn.net/?f=A%3D0.33%5C%20m%5E2)
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
![\dfrac{1}{2}\rho CAv_t^2=mg](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Crho%20CAv_t%5E2%3Dmg)
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
![C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01](https://tex.z-dn.net/?f=C%3D%5Cdfrac%7B2mg%7D%7B%5Crho%20Av_t%5E2%7D%5C%5C%5C%5CC%3D%5Cdfrac%7B2%5Ctimes%2075%5Ctimes%209.8%7D%7B1.225%5Ctimes%200.33%5Ctimes%20%2860%29%5E2%7D%5C%5C%5C%5CC%3D1.01)
So, the drag coefficient is 1.01.