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aalyn [17]
4 years ago
14

A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di

stance d1= 0.850 m to the left of sawhorse B. When the cat is a distance d2= 1.11 m to the right of sawhorse B, the plank just begins to tip.
If the cat has a mass of 2.9 kg, how far to the right of sawhorse B can it walk before the plank begins to tip?
Physics
1 answer:
dimulka [17.4K]4 years ago
6 0

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

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When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of
lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0
3 years ago
A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft
erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)

Now, differentiating both sides with respect to time 't', we get:

2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}

Now, we are given speed as 25 mph. So, \frac{de}{dt}=25\ mph

Also, when e=91\ mi, we can find 'x' using equation (1). This gives,

x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi

Now, plug in the values of 'e' and 'x' and solve for \frac{dx}{dt}. This gives,

\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0
3 years ago
A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is
kenny6666 [7]

Answer:

f2/f1 = \sqrt{2}

Explanation:

From frequency of oscillation

f = 1/2pi *\sqrt{k/m}

Initially with the suspended string, the above equation is correct for the relation, hence

f1 = 1/2pi *\sqrt{k/m}

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

f2 = 1/2pi *\sqrt{2k/m}

Employing f2/ f1, we have

f2/f1 = \sqrt{2}

3 0
3 years ago
What is the wavelength of a wave?
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I know for a fact the answer is D. the distance traveled by the wave during one full cycle
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