gravitational force of planet exerted on its object near the surface is known as weight
so here we know that gravitational force of mars is much less than the gravitational force of Earth
So on the surface of mars the Weight of objects must be much less than the weight of object on surface of Earth
so here correct answer must be
<em>D. Your weight would decrease.</em>
Answer:
109656.25 Nm
Explanation:
= Final angular velocity = 1.5 rad/s
= Initial angular velocity = 0
= Angular acceleration
t = Time taken = 6 s
m = Mass of disk = 29000 kg
r = Radius = 5.5 m
![\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha%20t%5C%5C%5CRightarrow%20%5Calpha%3D%5Cdfrac%7B%5Comega_f-%5Comega_i%7D%7Bt%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cdfrac%7B1.5-0%7D%7B6%7D%5C%5C%5CRightarrow%20%5Calpha%3D0.25%5C%20rad%2Fs%5E2)
Torque is given by
![\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm](https://tex.z-dn.net/?f=%5Ctau%3DI%5Calpha%5C%5C%5CRightarrow%20%5Ctau%3D%5Cdfrac%7B1%7D%7B2%7Dmr%5E2%5Calpha%5C%5C%5CRightarrow%20%5Ctau%3D%5Cdfrac%7B1%7D%7B2%7D29000%5Ctimes%205.5%5E2%5Ctimes%200.25%5C%5C%5CRightarrow%20%5Ctau%3D109656.25%5C%20Nm)
The torque specifications must be 109656.25 Nm
Answer:
66.4 N
Explanation:
From Newton's second law, <em>F </em>=<em> ma</em>
where <em>F</em> is the force, <em>m</em> is the mass and <em>a</em> is the acceleration.
Because the object has acceleration in two directions and the mass is constant, the force will be in two directions. The component of the forces are:
![F_x = ma_x = (7.00\text{ kg})(3.00 \text{ m/s}^2) = 21.0\text{ N}](https://tex.z-dn.net/?f=F_x%20%3D%20ma_x%20%3D%20%287.00%5Ctext%7B%20kg%7D%29%283.00%20%5Ctext%7B%20m%2Fs%7D%5E2%29%20%3D%2021.0%5Ctext%7B%20N%7D)
![F_y = ma_y = (7.00\text{ kg})(9.00 \text{ m/s}^2) = 63.0\text{ N}](https://tex.z-dn.net/?f=F_y%20%3D%20ma_y%20%3D%20%287.00%5Ctext%7B%20kg%7D%29%289.00%20%5Ctext%7B%20m%2Fs%7D%5E2%29%20%3D%2063.0%5Ctext%7B%20N%7D)
The magnitude of the resultant force is given by
![F = \sqrt{F_x^2+F_y^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7BF_x%5E2%2BF_y%5E2%7D)
![F = \sqrt{(21.0\text{ N})^2+(63.0\text{ N})^2} = \sqrt{(441.0\text{ N}^2)+(3969.0\text{ N}^2)} = \sqrt{(4410\text{ N}^2)} = 66.4 \text{ N}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7B%2821.0%5Ctext%7B%20N%7D%29%5E2%2B%2863.0%5Ctext%7B%20N%7D%29%5E2%7D%20%3D%20%5Csqrt%7B%28441.0%5Ctext%7B%20N%7D%5E2%29%2B%283969.0%5Ctext%7B%20N%7D%5E2%29%7D%20%3D%20%5Csqrt%7B%284410%5Ctext%7B%20N%7D%5E2%29%7D%20%3D%2066.4%20%5Ctext%7B%20N%7D)
The type of friction of a kite suspended
in the sky that is flowing back and forth is fluid friction. The fluid here is
the air that helps the kite move back and forth. The kite feels a drag force
due to air which acts in the upward direction.
the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.