Answer: A) mass on earth surface = 5.91kg
B) mass on surface of jupiter = 5.91kg
C) weight on surface of jupiter = 10.697N
Explanation:
The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below
W=mg
From the question, g= 9.8m/s² and weight on the surface on the earth is 58N
A) The mass of watermelon on earth is
m = 58/ 9.8 = 5.91kg
B) the mass of the watermelon on jupiter is 5.91kg.
You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.
C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as
W = mg
W = 5.91 x 9.8
= 10.697N.
An apple falling to the ground is not an example of centripetal acceleration.
Answer:![0.084 m/s^2](https://tex.z-dn.net/?f=0.084%20m%2Fs%5E2)
Explanation:
Given
Total time=27 min 43.6 s=1663.6 s
total distance=10 km
Initial distance ![d_1=8.13 km](https://tex.z-dn.net/?f=d_1%3D8.13%20km)
time taken=25 min =1500 s
initial speed ![v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7B8.13%5Ctimes%201000%7D%7B25%5Ctimes%2060%7D%3D5.6%20m%2Fs)
after 8.13 km mark steve started to accelerate
speed after 60 s
![v_2=v_1+at](https://tex.z-dn.net/?f=v_2%3Dv_1%2Bat)
![v_2=5.6+a\times 60](https://tex.z-dn.net/?f=v_2%3D5.6%2Ba%5Ctimes%2060)
distance traveled in 60 sec
![d_2=v_1\times 60+\frac{a60^2}{2}](https://tex.z-dn.net/?f=d_2%3Dv_1%5Ctimes%2060%2B%5Cfrac%7Ba60%5E2%7D%7B2%7D)
![d_2=336+1800 a](https://tex.z-dn.net/?f=d_2%3D336%2B1800%20a)
time taken in last part of journey
![t_3=1663.6-1560=103.6 s](https://tex.z-dn.net/?f=t_3%3D1663.6-1560%3D103.6%20s)
distance traveled in this time
![d_3=v_2\times t_3](https://tex.z-dn.net/?f=d_3%3Dv_2%5Ctimes%20t_3)
![d_3=\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=d_3%3D%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
and total distance![=d_1+d_2+d_3](https://tex.z-dn.net/?f=%3Dd_1%2Bd_2%2Bd_3)
![10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=10000%3D8.13%5Ctimes%201000%2B336%2B1800%20a%2B%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
![1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=1870%3D336%2B1800%20a%2B%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
![a=0.084 m/s^2](https://tex.z-dn.net/?f=a%3D0.084%20m%2Fs%5E2)
Answer:
Mechanical advantage = 15
Explanation:
Given the following data;
Output force = 3000N
Input force = 200N
To find the mechanical advantage;
Mechanical advantage = output force/input force
Substituting into the equation, we have
Mechanical advantage = 3000/200
Mechanical advantage = 15