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dangina [55]
2 years ago
15

A burning candle is covered by a jar as shown in the picture. The whole arrangement has a mass of 500 g. What will be the approx

imate mass of the arrangement when the candle is completely burnt, after four minutes?
A)
0 g



B)
50 g



C)
250 g



D)
500 g
Physics
2 answers:
topjm [15]2 years ago
7 0

Technically, I can't answer the question, because you won't
let me see the picture that goes along with it and is a part of it. 
But I'm familiar with the set-up, have dealt with the question before,
and I can answer it from my previous experience and general knowledge.

If there is 500g of mass inside the jar when you lower it over
the candle, then there will be 500g of mass at any time after that,
forever, or until you pick up the jar and take some mass out or put
some more in.  It doesn't matter how long you wait.  It also doesn't
matter whether or not the candle is burning, whether or not the sun
is shining on the jar, or whether somebody comes along and spray-paints
the outside of the jar with black paint.  Matter is not created or destroyed. 
Whatever mass was inside when the jar got closed stays in there.
 
weqwewe [10]2 years ago
7 0
The answer is D)<span>500 g  it will not change</span>
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Represent 8953 ms with Sl units having an appropriate prefix Express your answer to four significant figures and include the app
tatyana61 [14]

Answer:

8.953 s

Explanation:

Here a time of 8953 ms is given.

Some of the prefixes of the SI units are

mili = 10⁻³

micro = 10⁻⁶

nano = 10⁻⁹

kilo = 10³

Mega = 10⁶

Each prefix increases or decreases by a factor of 10³

The number is 8953.0

Here, the only solution where the number of significant figures is mili.  If any other prefix is chosen then the significant figures will increase.

1 milisecond = 1000 second

1\ second=\frac{1}{1000}\ milisecond

\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second

So, 8953 ms = 8.953 s

8 0
3 years ago
Why do astronauts appear to move in slow motion in space?
bazaltina [42]

Answer:

A few reasons. They are in low gravity environments, so they don't have the advantage of gravity to help move them.

Explanation:

6 0
3 years ago
Consider a force of 57.3 N, pulling 3 blocks of
andrezito [222]

Block 1 (the rightmost block) has

• net horizontal force

∑ <em>F</em> = <em>F</em> - <em>T₁</em> - <em>f₁</em> = <em>m₁a</em>

• net vertical force

∑ <em>F</em> = <em>N₁</em> - <em>m₁g</em> = 0

where <em>F</em> = 57.3 N, <em>T₁</em> is the tension in the string connecting blocks 1 and 2, <em>f₁</em> is the magnitude of kinetic friction felt by block 1, <em>m₁</em> = 0.8 kg is its mass, <em>a</em> is the acceleration you want to find, and <em>N₁</em> is the magnitude of the normal force exerted by the surface.

Block 2 (middle) has much the same information:

• net horiz. force

∑ <em>F</em> = <em>T₁</em> - <em>T₂</em> - <em>f₂</em> = <em>m₂a</em>

• net vert. force

∑ <em>F</em> = <em>N₂</em> - <em>m₂g</em> = 0

with similarly defined symbols.

The same goes for block 3 (leftmost):

• net horiz. force

∑ <em>F</em> = <em>T₂</em> - <em>f₃</em> = <em>m₃a</em>

• net vert. force

∑ <em>F</em> = <em>N₃</em> - <em>m₃g</em> = 0

We have <em>m₁</em> = <em>m₂</em> = <em>m₃</em> = 0.8 kg, so I'll just replace each with <em>m</em>. It follows that each normal force has the same magnitude, <em>N₁</em> = <em>N₂</em> = <em>N₃</em> = <em>mg</em>. And as a consequence of that, each frictional force has the same magnitude, <em>f₁</em> = <em>f₂</em> = <em>f₃</em> = 0.4<em>mg.</em>

In short, the relevant equations are

[1] … 57.3 N - <em>T₁</em> - 0.4<em>mg</em> = <em>ma</em>

[2] …<em>T₁</em> - <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>

[3] … <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>

<em />

Adding [1], [2] and [3] together eliminates the tension forces, and we get

57.3 N - 1.2<em>mg</em> = 3<em>ma</em>

<em />

Solve for <em>a</em> :

57.3 N - 1.2 (0.8 kg) (9.8 m/s²) = 3 (0.8 kg) <em>a</em>

57.3 N - 9.408 N = (2.4 kg) <em>a</em>

<em>a</em> = (47.892 N) / (2.4 kg)

<em>a</em> ≈ 20.0 m/s²

3 0
3 years ago
A stone is dropped from rest from the top of a building. It takes Δt = 2.2 s for it to reach the ground.
NeTakaya

Answer:

Value of magnitude of acceleration will be 9.8m/sec^2

Explanation:

It is given that when a stone is dropped it takes 2.2 sec to reach the ground

(a) As the stone is dropped from the top of building

So its initial velocity u_i will be 0 m /sec

(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity

So value of magnitude of acceleration will be equal to 9.8m/sec^2

6 0
3 years ago
A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje
Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

6 0
3 years ago
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