answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
vova2212 [387]

solution:

consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm

7 0
2 years ago
Bianca is standing at x =600m. Firecracker 1, at the origin, and firecracker 2, at x =900m, explode simultaneously. The flash fr
agasfer [191]

Answer:3 \mu s

Explanation:

Given

Bianca is at x=600 m

i.e. distance between origin and Bianca is 600 m

time taken to reach Bianca eyes is

t=\frac{600}{speed\ of\ light}

t=\frac{600}{3\times 10^8}

t=2\times 10^{-6} s

t=2 \mu s

i.e. Cracker exploded at t=2\mu s because it is observed at t=4\mu s

Time taken by second cracker flash to reach Bianca eyes

t_2=\frac{300}{3\times 10^8}

t_2=10^{-6}

t_2=1 \mu s

Therefore it will be observed at t=3 \mu s

4 0
2 years ago
A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons
liraira [26]

When the metals touch together, half the charge of the charged metal flows to the other because the electrons all repel each other. Therefore this also means that each metal ball contains the same amount of electrons. Each ball has 5^10 electrons, this is equivalent to a total charge of:

Q1 = Q2 = (1.602 * 10^-19 coulombs / electron) 5^10 electrons = Q

Q = 1.564 * 10^-12 C

 

Now using the Coulombs law to find for the electric force:

F = k q1 q2 / r^2 = k (Q)^2 / r^2

where k is a contant = 9 * 10^9 N m^2 / C^2

r = the distance of the two metals = 0.2 m

So,

F = (9 * 10^9 N m^2 / C^2) (1.564 * 10^-12 C)^2 / (0.2 m)^2

F = 5.51 * 10^-13 N

 

Since the two metals repel therefore they are the one which exerts the force hence the magnitude must be negative:

<span>F = - 5.51 * 10^-13 N</span>

7 0
2 years ago
A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu
Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 m/s^2

Explanation:

Mass (m) of the sailboat =  2000 kg

Force acting on the sailboat due to ocean  tide is F_1 = 3000N

Eastwards means takes place along the positive x direction

ThenF_{1x} = 3000N and F_{1y}= 0

Wind Force acting on the Sailboat isF_2  = 6000N directed towards the northwest that means at an angle  45 degree above the negative x axis

Then  

F_{2x} = -(6000N) cos 45 degree = -4242.6 N

F_{2y}  = (6000N) cos 45 degree = 4242.6 N

Hence  , the net force acting on the sailboat in x direction is  

F_x = F_{1x}+ F_{2x}

=  - 3000 N + 4242.6 N

=  - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is  

F_y = F_{1y}+ F_{2y}

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

\sqrt{(1242.6)^2 + (4242.6)^2

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

4420.8 N

F = ma

a = \frac{F}{m}

a =\frac{4420.8}{ 2000}

=2.2 m/s^2

4 0
2 years ago
Other questions:
  • Iron(II) carbonate (FeCO3) has a solubility product constant of 3.13 x 10-11 . Calculate the molar solubility of FeCO3 in water
    11·1 answer
  • Which of the following diagrams involves a virtual image ?
    9·1 answer
  • Compare the density, weight, mass, and volume of a pound of gold to a pound of iron on the surface of Earth.
    11·1 answer
  • A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
    14·1 answer
  • The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan
    14·1 answer
  • A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.
    11·1 answer
  • For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?
    10·1 answer
  • The arm of a crane at a construction site is 17.0 m long, and it makes an angle of 11.6 ◦ with the horizontal. Assume that the m
    7·1 answer
  • Rod A has twice the diameter of rod B, but both are made of iron and have the same initial length. Both rods are now subjected t
    11·1 answer
  • The solar energy strikes the deck at the rate of 1400 W on every square metre.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!