Here, we know, according to 3rd Equation of Kinematics,
v² - u² = 2as
Here, u = 0 [ Free fall ]
a = 9.8 m/s² [ constant value for the Earth system ]
s = 15 m
Substitute their values,
v² - 0² = 2 * 9.8 * 15
v² = 294
v = √294
v = 17.15 m/s
In short, Your Answer would be Option D
Hope this helps!
Answer:
1,800kg
Explanation:
Force = Mass x Acceleration
F = m x a
15840N = m x 8.8
8.8 x m = 15840
m = 15840/8.8
= 1,800kg
Answer:
Yes
Explanation:
The variables change in and experiment.
Answer:
Explanation:
a ) Entropy change dS = dQ/T
= mcdT /T
Integrating both sides
S₂ - S₁ = - mclnT₂ /T₁
= - .5 X 4200 ln (85+273) /( 20 + 273 )
.5 X 4200 ln 358/ 293
= - 417.6 J/K
Entropy change will be negative as heat is lost by the system .
b ) Sine there is no change in the temperature of air , This heat will enter air at temperature ( 20+ 273) K = 293 K
Heat entering air
= .5 x 4200 x 65
= 136500 J
Change in entropy
136500 / 293 ( room temperature is constant at 293k
= + 465.87 J/K
Entropy change will be positive as heat is gained by the system .
Total change in the entropy of the system (tea + air )
= +465.87 - 417.6
= 48.27 J/K
Entropy change will be negative as heat is lost by the system .
The magnitude of the frictional force of the box is 65 N
From the question given above, the following data were obtained:
- Mass (m) = 25 Kg
- Force applied (Fₐ) = 135 N
- Net force (Fₙ) = 70 N
- Frictional force (Fբ) =?
The frictional force acting on the box can be obtained as follow:
Fₙ = Fₐ – Fբ
70 = 135 – Fբ
Collect like terms
70 – 135 = – Fբ
–65 = –Fբ
Multiply through by –1
<h3>Fբ = 65 N</h3>
Thus, the frictional force of the box is 65 N
Learn more on Frictional force: brainly.com/question/25767576
