<h2>
Mass of object in Earth is 1.37 kg</h2>
Explanation:
On planet B where the magnitude of the free-fall acceleration is 1.91g , the object weighs 25.74 N.
We have
Weight = Mass x Acceleration due to gravity
On planet B
25.74 = Mass x 1.91 g
25.74 = Mass x 1.91 x 9.81
Mass = 1.37 kg
Mass is constant for an object. It will not change with location.
Mass of object in Earth = Mass of object in Planet B
Mass of object in Earth = 1.37 kg
First solve the potential energy of the biker. using the fomula:
PE = mgh
where m is the mass of the object
g is the acceleration due to gravity ( 9.81 m/s2)
h is the height
PE = 96 kg ( 1120 m ) ( 9.81 m/s2)
PE = 1054771.2 J
then power = Work / time
P = 1054771.2 J / ( 120 min ) ( 60 s / 1 min)
P = 146.5 W
The actual answer is 165 miles, but using significant figure rules the answer is 200. This is because the sig fig rules are as follows ...
<span>1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
<span>3. A final zero or trailing zeros in the decimal portion ONLY are significant.
</span></span>
So the zeroes in a number like 20 or 23,000 are NOT significant. When you add numbers you must find the addend with the lowest amount of significant figures and round the answer to that. In this case most of the addends only have one sig fig, so you round 165 to 200 to make it only have one sig fig.
Moment about D=0
or,T1*6a-5w*3a-9w*2a=0
T2+T1=5w+9w
Answer:
Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e 
= 875 kg.m²
We know from the conservation of angular momentum that:
![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
From the conservation of energy as well;we have :
^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)
