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Lyrx [107]
3 years ago
9

A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig

ht triangles. At what rate is the length of the person's shadow changing when the person is 16 ft from the lamppost?
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

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andrezito [222]
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6 0
3 years ago
Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?
frozen [14]

Answer:

According to super hero logic , nothing will happen to him.

But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.

He may / may not be affected but his suit will be damaged for sure as it is made of metal.

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8 0
4 years ago
The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms
aalyn [17]

The question to the above information is;

What is the best use of an atomic model to explain the charge of the particles in Thomson's beams?

Answer;

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation;

-Atoms are comprised of a nucleus consisting of protons (red) and neutrons (blue). The number of orbiting electrons is the same as the number of protons and is termed the "atomic number" of the element.

J.J. Thomson discovered the electron. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

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5 0
3 years ago
Read 2 more answers
The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =
dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar,  θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

                        ∑ Mₐ = Iα

Where,     Mₐ - the moment about A

                 α  - angular acceleration

                 I - moment of inertia of the rod AB

                       -mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha

                        \alpha=\frac{-3gcos\theta}{2l}

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

                          \vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

The acceleration at the center of the bar

                          \vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

                         (\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]

Comparing the coefficients of i

=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}

Comparing coefficients of j

(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}

Net force on x direction

F_{x}=(\vec{a_{G}})_{x}

substituting the values

F_{x}=1.5(14.58L+11.96)

Similarly net force on y direction

F_{y}=(\vec{a_{G}})_{y}+mg

               = 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

F=\sqrt{F_{x}^{2}+F_{y}^{2}}

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0
3 years ago
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
3 years ago
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