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3 years ago

14

How much time does it take for tweety’s bird cage to hit the ground after it was dropped if it reached a velocity of 22 meters p

er second right before impact?

1 answer:

guajiro [1.7K]3 years ago

7 0

It matters on the weight

You might be interested in

What is the correct answer?

sashaice [31]

Answer:

D) 4

Explanation:

Roots of a polynomial must be factors of the last term.

In this case, the factors of 6 are +1, -1, +2, -2, +3, -3, +6, -6. The only factor that doesn't show up, given the options, is 4. This means that D is the correct answer.

7 0

3 years ago

A student is throwing a ball. She throws the ball up with her left hand, it passes over her head, and she catches it with her ri

pychu [463]

Answer:

(c) at point 2, the ball is at its highest height do its PE is max. Also at ms height, velocity is zero therefore KE is zero.

8 0

2 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 !

$\texttt{ }$

<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>

<h3>Step A:</h3>

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

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

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

$\texttt{ }$

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

$W_A = -P_2(V_2 - V_1)$

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

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

$\texttt{ }$

<h3>Step B:</h3>

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

$P_2V_2 = P_3V_3$

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

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

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

$W_B = -P_3(V_3 - V_2)$

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

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

$\texttt{ }$

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

$W = W_A + W_B$

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

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

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

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

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

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

$\texttt{ }$

<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

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?

iren2701 [21]

Answer:

$K.E=29.403125J$

Explanation:

From the question we are told that

Mass $M=100$

Height $50-20=30m$

Generally the equation for velocity before impact is is is mathematically given by

$v=\sqrt{2gh}$

$v=\sqrt{2*9.8*30}$

$v=24.25$

Generally the equation for Kinetic Energy is is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*100*(24.25)^2\\$

$K.E=29403.125J$

$K.E=29.403125J$

8 0

3 years ago

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

8 0

3 years ago