Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener
1 answer:
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Answer:
≈ 3077.34
Explanation:
To calculate the flux of F (vector field) across surface S, where
F(x,y,z) = y i − x j +and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π
We will evaluate the following integral:
Substituting for the surface
x = u cos v
y = u sin v
z = v
Then
F(S(u,v)) = u sin v i - u cos v j + k
The normal vector N is computed as
Where:
N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v
N = < 2v sin v, -2v cos v, u >
F(S(u,v)).N = < u sin v, -u cos v, >. < 2v sin v, -2v cos v, u >
F(S(u,v)).N = 2uv + u
Thus
≈ 3077.34
Answer:
1/7 kg
Explanation:
Refer to the attached diagram for enhanced clarity regarding the question.
One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.
g denotes the acceleration due to gravity.
Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.
Given M = 1.0 kg and a = 3/4g.
By applying Newton's second law;
For the body with mass m;
T - mg = ma... (1)
For the body with mass M;
Mg - T = Ma... (2)
Combining equations 1 and 2 gives;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into this equation leads to;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
Multiplying gives: 28m = 4
m = 1/7 kg
Hence, the mass of the other box is 1/7 kg
Answer:
Explanation:
The first number is .
The second number is .
We must multiply these two numbers together.
In scientific notation:
Therefore, this is the solution you are looking for.