Answer:
The mass of the object is approximately 70.79 kilograms
Explanation:
We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:
![F_{net}=m\,a](https://tex.z-dn.net/?f=F_%7Bnet%7D%3Dm%5C%2Ca)
Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:
Answer:
+ 0.07 C
Explanation:
From the question given above, the following data were obtained:
Potential difference (V) = 12 V
Energy (E) = 0.418 J
Charge (Q) =?
The energy (E) , potential difference (V) and charge (Q) are related by the following equation:
E = ½QV
With the above formula, we can obtain the charge as follow:
Potential difference (V) = 12 V
Energy (E) = 0.418 J
Charge (Q) =?
E = ½QV
0.418 = ½ × Q × 12
0.418 = Q × 6
Divide both side by 6
Q = 0.418 / 6
Q = + 0.07 C
Answer:
6 seconds
Explanation:
![d=v_ot+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=d%3Dv_ot%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Since the object is being dropped out of the window, it has no initial velocity.
![176.4=\dfrac{1}{2}(9.8)t^2 \\\\t^2=36 \\\\t=6\text{ seconds}](https://tex.z-dn.net/?f=176.4%3D%5Cdfrac%7B1%7D%7B2%7D%289.8%29t%5E2%20%5C%5C%5C%5Ct%5E2%3D36%20%5C%5C%5C%5Ct%3D6%5Ctext%7B%20seconds%7D)
Hope this helps!
Answer:
42.69 N and 18.07 N
Explanation:
We are given that
Mass of ladder=6.2 kg
Length of ladder=1.97 m
Distance of Sawhorse A from one end=0.64 m
Distance of sawhorse B from other end=0.17 m
Let center of Ladder=![\frac{1.97}{2}=0.985 m](https://tex.z-dn.net/?f=%5Cfrac%7B1.97%7D%7B2%7D%3D0.985%20m)
Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m
Distance of sawhorse B from center of ladder=0.985-0.17=0.815 m
Force one ladder due to gravity=mg=![6.2\times 9.8=60.76N](https://tex.z-dn.net/?f=6.2%5Ctimes%209.8%3D60.76N)
Where ![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
Torque applied on Sawhorse A=![0.345F_a](https://tex.z-dn.net/?f=0.345F_a)
Torque applied on Sawhorse B=![0.815F_b](https://tex.z-dn.net/?f=0.815F_b)
In equilibrium
![0.345F_a=0.815F_b](https://tex.z-dn.net/?f=0.345F_a%3D0.815F_b)
![F_b=\frac{0.345}{0.815}F_a](https://tex.z-dn.net/?f=F_b%3D%5Cfrac%7B0.345%7D%7B0.815%7DF_a)
Total force=![F_a+F_b](https://tex.z-dn.net/?f=F_a%2BF_b)
![F_a+\frac{0.345}{0.815}F_a=60.76](https://tex.z-dn.net/?f=F_a%2B%5Cfrac%7B0.345%7D%7B0.815%7DF_a%3D60.76)
![\frac{0.815F_a+0.345F_a}{0.815}=60.76](https://tex.z-dn.net/?f=%5Cfrac%7B0.815F_a%2B0.345F_a%7D%7B0.815%7D%3D60.76)
![\frac{1.16}{0.815}F_a=60.76](https://tex.z-dn.net/?f=%5Cfrac%7B1.16%7D%7B0.815%7DF_a%3D60.76)
![F_a=\frac{60.76\times 0.815}{1.16}=42.69 N](https://tex.z-dn.net/?f=F_a%3D%5Cfrac%7B60.76%5Ctimes%200.815%7D%7B1.16%7D%3D42.69%20N)
![F_b=\frac{0.345}{0.815}\times 42.69=18.07 N](https://tex.z-dn.net/?f=F_b%3D%5Cfrac%7B0.345%7D%7B0.815%7D%5Ctimes%2042.69%3D18.07%20N)