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Taya2010 [7]
2 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:
algol132 years ago
4 0

Answer:

Explanation:

Find attached the solution

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A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°
Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume V_{1} = 2.00\ L

Pressure P_{1}= 1.00\ atm

Pressure P_{2}= 70.0\ kPa

Temperature T_{1}= 20.0°C = 293\ K

Temperature T_{2}= 7.00°C = 280\ K

We need to calculate the volume at mountains

Using  gas law

\dfrac{PV}{T}=\ Constant

For both temperature,

\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}

Put the value into the formula

\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}

V_{2}=\dfrac{101.325\times2\times280}{293\times70}

V_{2}=2.766\ litre

Hence, The volume at mountains is 2.766 L.

5 0
1 year ago
Based on the emf value measured at frame 700, what is the average magnitude of the magnetic field inside the magnet assembly? No
Nadusha1986 [10]

Answer:

The average magnitude of magnetic field B= 0.0433/ d Tesla

(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)

Explanation:

Induced emf

where B= magnetic field  

d= breadth of rectangular piece

V= velocity with which the rectangular piece = o.o6m/s

n= no of turns  = 10

EMF = 26mV

since d (breadth of the frame) is not given, I will use it as a variable

EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)

From eq 1, we get

B= (EMF)/(n d V)

B= (26 X 0.001) / (10 d 0.06)

B= 0.0433/ d Tesla

4 0
1 year ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
1 year ago
An object weighs 200 newtons at a distance of 100 kilometers above the center of a small uniform planet. how much will the objec
disa [49]

Since the law of gravitation is an inverse square law if you quadruple the radius the f will drop by a factor of 16 SO the object would weigh 200/16 = 12.5N

In other words, as the distance, or radius, quadruples the weight becomes 1/16 of the original weight. Just plug in 4 for r and when you square it you get 16. The numerator is 1 so that is how the weight becomes 1/16.

7 0
2 years ago
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