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hodyreva [135]
4 years ago
12

A Martian rover is descending a hill sloped at 30° with the horizontal. It travels with a constant velocity of 2 meters/second.

Calculate its horizontal velocity.
Physics
2 answers:
kodGreya [7K]4 years ago
8 0
Vx = 2*cos30 = 1.73m's
Other words:
D = Vo*t = 1.4 * 21 = 31.5m. @ 30 Deg.
Dy = 31.5*sin30 = 15.75 m.
uysha [10]4 years ago
7 0

<u>Answer:</u> The horizontal velocity of rover is 1.73 m/s

<u>Explanation:</u>

Horizontal velocity is defined as the velocity of body proceeding in the horizontal direction. This velocity proceeds in the 'x' direction.

Mathematically,

V_x=V_o\cos \theta

where,

V_o = constant velocity of rover = 2 m/s

\cos \theta = cosine function of angle 30°

Putting values in above equation, we get:

V_x=2\times \cos (30^o)\\\\V_x=1.73m/s

Hence, the horizontal velocity of rover is 1.73 m/s

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Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

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A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

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Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

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3 0
3 years ago
The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the
Ivanshal [37]

Answer:

(a) \alpha = - 1.32\ rev/m^{2}

(b) \theta = 13674\ rev

(c) \alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}

(d) a = 22.458\ m/s^{2}

Solution:

As per the question:

Angular velocity, \omega = 190\ rev/min

Time taken by the wheel to stop, t = 2.4 h = 2.4\times 60 = 144\ min

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

\omega' = \omega + \alpha t

omega' = final angular velocity

\omega = initial angular velocity

\alpha = angular acceleration

Now,

0 = 190 + \alpha \times 144

\alpha = - 1.32\ rev/m^{2}

Now,

(b) The no. of revolutions is given by:

\omega'^{2} = \omega^{2} + 2\alpha \theta

0 = 190^{2} + 2\times (- 1.32) \theta

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(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}

(d) The radial acceleration is given by:

\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s

Linear acceleration is given by:

a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}

a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}

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The speed of the sound wave in the medium, given the data is 3900 m

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<h3>How to determine speed of the sound wave</h3>

The speed of the wave can be obtained as illustrated below:

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