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ZanzabumX [31]
3 years ago
13

Convert

n)}{L}" alt="\frac{(0.779mg)(min)}{L}" align="absmiddle" class="latex-formula"> into \frac{(kg)(s)}{m^{3} }
Please help. I am getting multiple different answers.
Physics
1 answer:
Orlov [11]3 years ago
5 0

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

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3 years ago
a body is dropped from a height of 240m. (i)How long does it take to reach the ground. (ii) What is the velocity with which it w
kramer

Answer:

7.0 s, 69 m/s

Explanation:

If we take down to be positive, then the time to reach the ground is:

x = x₀ + v₀ t + ½ at²

240 m = (0 m) + (0 m/s) t + ½ (9.8 m/s²) t²

t = 7.0 seconds

The final velocity is:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²) (240 m - 0 m)

v = 69 m/s

5 0
3 years ago
In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b
V125BC [204]
The total average velocity v=+1.41 m/s (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion (S_1=6.30 km=6300 m, due west) and of the displacement in the second part of the motion (S_2, due east).
-The total time taken t is the time taken for the first part of the motion, t_1, and the time taken for the second part of the motion, t_2. t_1 can be found by using the average velocity and the displacement of the first part:
t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s

t_2, instead, can be written as \frac{S_2}{v_2}, where v_2=-0.630 m/s is the average velocity of the second part of the motion (with a negative sign, since it is due east). 

Therefore, we can rewrite the initial equation as:
v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
S_2=-844 m=-0.844 km

4 0
3 years ago
Use the general formulas for gravitational force and centripetal force to derive the relationship between speed (v) and orbital
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Solution :

We know that :

Formula for Gravitational force is given by :

$F_g=\frac{Gmn}{r^2}$

where, G is the gravitational constant

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            r  is the distance between the two bodies.

And the formula for the centripetal force is given by :

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where, m is the mass of the rotating body

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We know that mathematically, the gravitational force is equal to the centripetal force of the body.

Therefore,

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800,000 km/hr is the answr
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