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3 years ago

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A driver fills an 18.9L steel gasoline can with gasoline at 15.0°C right up to the top. He forgets to replace the cap and leaves

the can in the back of his truck. The temperature climbs to 30.0°C by 1pm. How much gasoline spills out of the can

1 answer:

GarryVolchara [31]3 years ago

5 0

Answer:

ΔV = 0.98 L

Explanation:

First, we will calculate the increased volume using Charles' Law:

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where,

V₁ =initial volume = 18.9 L

V₂ = final volume = ?

T₁ = initial temperature = 15°C + 273 = 288 k

T₂ = final temperature = 30°C + 273 = 303 k

Therefore,

$\frac{18.9\ L}{288\ k} = \frac{V_2}{303\ k}$

V₂ = 19.88 L

Now, we calculate the change in volume:

ΔV = V₂ - V₁ = 19.88L - 18.9 L

<u>ΔV = 0.98 L</u>

This is the volume of gasoline that will spill out.

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Provide one example of this law during a physical change.

Sergio [31]

Answer:

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Explanation:

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3 years ago

Se tiene un objeto a una altura de 8 m generando 450 J de energía potencial cuál es la masa de dicho objeto

fgiga [73]

Answer:

5.73kg

Explanation:

Given data

Height = 8m

Potential Energy = 450J

Required

The mass of the object

From the following expression

PE=mgh

we can substitute g= 9.81m/s^2 and the other parameters to get mass m

450= m*9.81*8

450= m* 78.48

m= 450/78.48

m= 5.73kg

Hence the mass is 5.73kg

8 0

3 years ago

A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same tim

Vikentia [17]

Answer:

(a) Friction force = 50 N

(b) Work done by friction = 300 j

(c) Net work done = 0 j

Explanation:

We have given that the box is pulled by 6 meter so d = 6 m

Force applied on the box F = 60 N

We have have given that velocity is constant so acceleration will be zero

So to applied force will be utilized in balancing the friction force

So friction force $F_{friction}=50N$

Work done by friction force $W_{friction}=F_{friction}\times d=50\times 6=300j$

Work done by applied force $W=F\times d=50\times 6=300j$

So net work done = 300-300 = 0 j

7 0

3 years ago

An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

So

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

So

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

8 0

3 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

3 years ago