What happens to the Total Energy as the spring bounce?

Answer it asap<br> I promise i will mark them the Brainly

nataly862011 [7]

Answer:

A) The acceleration is zero

<em>B) The total distance is 112 m</em>

Explanation:

<u>Velocity vs Time Graph</u>

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.

A. To calculate the acceleration, we use the formula:

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

$\displaystyle a=\frac{8-8}{12-0}=0$

This confirms the previous conclusion.

B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.

Area of rectangle= base*height=12 s * 8 m/s = 96 m

Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m

The total distance is: 96 m + 16 m = 112 m

4 0

3 years ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

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<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

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<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

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$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

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<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass of 1.56 kg. Calculate it

Kazeer [188]

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = $\frac{m}{V}$

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = $\frac{1,56Kg}{1,767145L}$ = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) / φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

3 0

3 years ago

Which of the following is an example of balanced forces?

Sergio [31]

A seesaw remains stationary when two students of equal weight sit on the ends

c

6 0

3 years ago

Read 2 more answers

What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6100 n

Hoochie [10]

Force on electron due to electric field is given by

$F = eE$

$F = 1.6 * 10^{-19}* 6100$

$F = 9.76 * 10^{-16} N$

now the acceleration is given by

$a = \frac{F}{m}$

$a = \frac{9.76 * 10^{-16}}{9.1 * 10^{-31}}$

$a = 1.07 * 10^{15} m/s^2$

so above is the magnitude of acceleration and its direction is opposite to field as electron is negatively charged so direction is towards SOUTH

4 0

3 years ago