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dimaraw [331]
3 years ago
6

A bin has a volume of 1.5m^3, what is its volume in ft^3?

Physics
1 answer:
il63 [147K]3 years ago
6 0

Answer:

1.5m^3=1.5\times 3.28^3ft^3=52.931ft^3

Explanation:

We have given volume V = 1.5m^3

We have to convert this volume into ft^3

We know that 1 m = 3.28 feet

So for converting  1.5m^3  to ft^3 we have to multiply 3.28^3 to 1.5

So 1.5m^3=1.5\times 3.28^3ft^3=52.931ft^3

So in cubic meter 1.5 will be equal to 82.923 cubic feet

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Please help on this one?
Lelu [443]

The easiest way to answer this question is just to get the answer first. The answer is A with the added comment that no chemical reaction has taken place.

Layered means that the chemicals are not soluble in one another. B is not the answer.

C is eliminated by what what was said about A.

D a solution is not a pure substance (singular) by itself. There are at least 2 chemicals together.

6 0
3 years ago
If the potential difference across the bulb in a camping lantern is 9.0 V, what is the
Gre4nikov [31]

Answer:

9V

Explanation:

The potential difference across the terminal as the same and thats because we are assuming that the source has no internal resistance.

Internal resistance are usually little resistances in the supply.

4 0
3 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
3 years ago
On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
sesenic [268]

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

7 0
3 years ago
If anyone anwers this question, i will give you 100 points and brainliest!
Tju [1.3M]

Answer:

33. at the top?

34. to add more "energy" to the water and keep it moving

35. the dam because the energy is is building up

Explanation:

i tink that's correct sorry if not

have a good rest of ur day

3 0
2 years ago
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