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Taya2010 [7]
3 years ago
5

Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coy

ote steps onto one end and pivots around its center of mass. Roadrunner is standing on the other end, which is held up by a small piece of ground underneath. The lever has mass 1.2 kg and length 16 m, the Roadrunner has mass 3.2 kg, and there is no friction at the pivot point of the lever.
a) Draw free body diagrams for the lever, Coyote, and Roadrunner. Make sure to label where all of the forces act on the lever’s free body diagram.
b) What is the minimal mass that Coyote must have to cause the lever to start rotating?
c) The coyote actually has a mass of 18 kg. If the roadrunner goes flying when the lever gets to a 45∘ angle from the horizontal, what is the roadrunner’s speed at that time, assuming the pivot is frictionless?
d) (Optional/challenge) What is the angular acceleration just after Coyote steps onto the lever?

Physics
1 answer:
algol133 years ago
4 0

Answer:

Explanation:

Find attached the solution

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Two small identical conducting spheres are placed with their centers 0.41 m apart. One is given a charge of 12 ✕ 10−9 C, the oth
nataly862011 [7]

(a) -1.48\cdot 10^{-5}N

The electrostatic force exerted between the two sphere is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the charges on the two spheres

r is the separation between the centres of the two spheres

In this problem,

q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m

Substituting these values into the equation, we find the force

F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

(b) +1.62\cdot 10^{-6}N

The total net charge over the two sphere is:

Q=q_1 +q_2 = 12\cdot 10^{-9}C+(-23\cdot 10^{-9}C)=-11\cdot 10^{-9} C

When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C

So the electrostatic force between the two spheres will now be

F=k\frac{q^2}{r^2}

And substituting numbers, we find

F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N

and the positive sign means the force is repulsive, since the two spheres have same sign charges.

7 0
3 years ago
P1: Explain in detail - How can the motion of an object that is already moving change? (What are the different ways a moving obj
Tomtit [17]

Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

5 0
4 years ago
What three sprites are needed to create the Space Invaders game that was discussed in the unit? spaceship, laser, enemy gnome, a
Lilit [14]

Answer: anlien, enemy gnome, spaceship

Explanation:

3 0
2 years ago
Which statement correctly explains how the Hertzsprung-Russell diagram helps to compare stars?
tekilochka [14]

Answer:

The Hertzsprung-Russell diagram plots stars according to their luminosity and temperature, which is also associated with spectral class.​

Explanation:

The Hertzsprung-Russell is a diagram representing:

- The surface temperature of the stars on the x-axis

- The absolute luminosity of the stars on the y-axis

So, the H-R diagram shows the relationship between the luminosity (power output) of a star and its temperature.

Most of the stars in the diagram are located along a diagonal line going from the top left to the bottom right: the stars in this diagonal are the stars in the main sequence, which is the main phase of life of a star.

Also, the graph shows that for stars in the main sequence, as the temperature increases, the luminosity also increases: these are the blue supergiant stars located in the top left corner.

Then, apart from the main sequence line, there are other groups of stars located in the top right corner (red giants) and in the bottom left corner (red, white dwarfs).

So, the correct answer is

The Hertzsprung-Russell diagram plots stars according to their luminosity and temperature, which is also associated with spectral class.​

4 0
4 years ago
A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m
Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

     6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

     2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0   (1)

2.070k + Φ = 70.53.    (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0
4 years ago
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