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3 years ago

Wiley The Coyote problem Pls Help Physics

Physics

1 answer:

antoniya [11.8K]3 years ago

3 0

Explanation:

Find the time it takes for the roadrunner to land.

Given (in the y direction):

Δy = 0 m

v₀ = v sin 10°

a = -9.81 m/s²

Find: t as a function of v

Δy = v₀ t + ½ at²

(0 m) = (v sin 10°) t + ½ (-9.81 m/s²) t²

t = (v sin 10°) / 4.905

Given (in the x direction):

Δx = 20.5 m

v₀ = v cos 10°

a = 0 m/s²

Find: t as a function of v

Δx = v₀ t + ½ at²

(20.5 m) = (v cos 10°) t + ½ (0 m/s²) t²

t = 20.5 / (v cos 10°)

Set equal and solve for v:

(v sin 10°) / 4.905 = 20.5 / (v cos 10°)

v² sin 10° cos 10° = 100.5525

v = 24.2485398301588

Graph:

desmos.com/calculator/x4b2zf1hxj

None of the options shown are correct.

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A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

3 years ago

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

3 years ago

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A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a

sladkih [1.3K]

Answer:

$d=1.29*10^{-6}m$

Explanation:

From the question we are told that:

Distance of wall from CD $D=1.4$

Second bright fringe $y_2= 0.803 m$

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

$y=frac{n*\lambda*D}{d}$

Where

$d=\frac{n*\lambda*D}{y}$

$d=\frac{2*431 *10^{-9}m*1.4}{0.803}$

$d=1.29*10^{-6}m$

8 0

3 years ago

What is the charge of an atom with 21 protons and 6 electrons

Svetach [21]

Answer:

Scandium with an ion charge of +3

Explanation:

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What type of circuit is illustrated?

Reil [10]

Answer:

I beleive its B

Explanation:

If not then A but I'm positive its B

6 0

3 years ago