Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:

By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
Answer:
B. x - t graph
Explanation:
A position-time (x-t) graph is a graph of the position of an object against (versus) time.
Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.
An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.
Hence, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.
<u>For example;</u>
Given that the equation of motion is S(t) = 4t² + 2t + 10. Find the instantaneous velocity at t = 5 seconds.
Solution.
Differentiating the equation, we have;
Substituting the value of "t" into the equation, we have;
S(5) = 42 m/s.
Answer:
I think u are traeling at speed of light and not ur friend
Explanation:
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.
Answer:
m = 375 [gram]
Explanation:
A triple Beam balance is an instrument very easy to use, since we only have to perform the arithmetic sum of each of the weights that are recorded in each beam
m = 300 + 70 + 5 = 375 [gram]
For a better understanding, the following image is attached, with values on each beam, which should be read.
The largest mass is in the indicator of 100 [gram], the second mass is in the indicator of 20 [gram] and the third is in the indicator of 5.8 [gram]. Thus the arithmetic sum corresponds to:
M= 100 + 20 + 5.8 = 125.8 [gram]
Note: it is important that when the instrument is in balance, the opposite end of the beam should indicate a position of zeros.