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3 years ago

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MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle

of 60.° at this speed, what is its maximum height? b) What would the launch angle have to be to reach a height of 46. cm?

2 answers:

labwork [276]3 years ago

5 0

Answer:

(a) $y_{max}=0.423m$

(b) $\alpha =64.3^{o}$

Explanation:

Given data

$v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle$

Solution

For Part (a)

As the velocity component in direction of y is given by:

$v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s$

The maximum displacement is given by:

$v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m$

For Part (b)

To reach y=46cm =0.46m apply:

$0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}$

Aleksandr [31]3 years ago

4 0

Explanation:

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Soloha48 [4]

Answer:

The mother has to sit 2.17 ft from the center on the other side of the seesaw.

Explanation:

We are trying to find the sum of torques given by the weights of mother and daughter to be zero.

If the torque of the daughter on one side of the pivoting point is given by:

5.5 ft x 63.5 lb x g = 349.25 g ft lb

we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:

d x 160.9 lb x g = 349.25 g ft lb

d = 2.17 ft

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3 years ago

What is moment? Write down the law of moment.A long spanner is used to unscrew the tight nut.Why?

vfiekz [6]

Answer:

Moment is the product of force and its perpendicular distance from a point along its line of action.

The net moments on clockwise and anticlockwise is zero at a point.

The net force applied on a spanner handle is equal to the moment of the rotation

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3 years ago

In your research lab, a very thin, flat piece of glass with refractive index 1.50 and uniform thickness covers the opening of a

lara31 [8.8K]

Answer:

The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ = 3472.

Explanation:

Refractive index of Glass (given) = 1.5

For the case of a constructive interference,

2nt = (m + 1/2) λ

For case 1,

2nt = (m + 1/2) 496 nm

For case 2,

2nt = (m +1+ 1/2) 386 nm

2nt = (m+3/2) * 386 nm

(m + 1/2) 496 nm = (m+3/2) * 386 nm

m = 3

Inserting the value of m in 1.

2nt = (m + 1/2) 496 nm

2*1.5t = (3 + 1/2) * 496 nm

t = ((3 + 1/2) * 496 nm)/ 3

t = 578.6 nm

The thickness of the glass, t = 578.6 nm

b)

It is generally known that for constructive interference,

2nt = (m + 1/2) λ

λ = 2nt / ((m + 1/2))

For Longest Wavelength, m = 0

λ = 2*1.5*578.6/ (1/2)

λ = 3472 nm

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3 years ago

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

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3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

4 years ago