Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²
Answer:
The first option, "The more mass an object has, the greater the gravitational pull.
"
Explanation:
Newton's Law of Gravity states that 
, where G is the gravitational constant, 
and 
are the masses of the two objects, and <em>r</em> is the distance between the objects' centers. Because the objects' masses are in the fraction's numerator, they are directly proportional to 
, and increasing the mass of one or both objects increases the gravitational pull.
Please have a great day! I hope this helps you understand the question!
0.600
In a regression analysis, if sse = 200 and ssr = 300, then the coefficient of determination is: <u>0.600</u>
<u></u>
1. You have that the data set having SSR=300 and SSE=200
2. Therefore you have the coefficient of determination is:
r²=SSR/SSTO
SSTO=SSR+SSE
3. Then, when you substitute the values, you obtain:
SSTO=200+300
SSTO=500
r²=300/500
4. So, you have that the result is:
r²=0.6
As a result, as you can see, the solution to the previous exercise is that the coefficient of determination is 0.6
<h3>What exactly is the connection between SST, SSR, and SSE?</h3>
- The difference between SST and SSR is the amount of Y's variability that is still unaccounted for after using the regression model, also known as the sum of squared errors (SSE). Sum of squares of residual can be used to directly calculate SSE.
To learn more about regression analysis visit:
brainly.com/question/12213669
#SPJ4
Http://www.sengpielaudio.com/calculator-ohm.htm
Answer:a=v-u/t
=23-8/3
=5m/s hope you got your answer
Explanation:
