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dangina [55]
3 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]3 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]3 years ago
7 0
The answer is D)<span>500 g  it will not change</span>
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Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi
max2010maxim [7]

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact  momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

3 0
3 years ago
A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in
Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

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The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

\frac{GMm}{d^{2}}=\frac{mv^2}{d}

v=\sqrt{\frac{GM}{d}}

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}

v = 1.81 x 10^-4 m/s

7 0
3 years ago
Find expressions for the force needed to bring an object of mass m from rest to speed v in time t. express your answer in terms
VARVARA [1.3K]
Good morning.

We have that:

\mathsf{V = a\cdot t} , since we have rest in the inicial time.

The acceleration can be found with Newton's Law:

\mathsf{F = m\cdot a\iff a = \dfrac{F}{m}}

Now we put the acceleratin in the velocity equation:

\mathsf{V = \dfrac{F}{m} \cdot t}

We want the force, so, let's isolate F:

\mathsf{V\cdot m = F\cdot t}\\ \\ \\ \boxed{\mathsf{F = \dfrac{V\cdot m}{t}}}

3 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb
Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

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This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

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Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

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This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

⇒ The force applied on one wheel during braking.

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