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Mama L [17]
4 years ago
14

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thr

ust) of 3432 N, the craft descends at constant speed; if the engine provides only 2317 N, the craft accelerates downward at 0.39 m/s2. What is the weight of the landing craft in the vicinity of Callisto's surface?
Physics
1 answer:
kolbaska11 [484]4 years ago
8 0

Answer:

W = mg = 3432 N

Explanation:

The force applied by the engine, F = 3432 N (upward) the craft descends at constant speed.

If the engine provides only 2317 N, the craft accelerates downward at 0.39 m/s².

On applying the force of 3432 N, the ship descends at a constant velocity such that the net force acting on it equal to 0. So, the upward normal force is equal to the weight of the landing craft.

So, the weight of the landing craft in the vicinity of Callisto's surface is 3432 N. Hence, this is the required solution.

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How does energy travel in a mechanical wave?
VLD [36.1K]

Answer: in a medium

Explanation:

Mechanical waves require medium to transfer energy. The medium particles vibrate about their fixed position to transfer energy.  There is no transport of medium particles.

Example, sound waves are mechanical waves. The sound travels through compression and rarefaction of the medium particles.

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4 years ago
1. A meteorologist describes a tropical storm as traveling northwest at 50 mi/h. Which attribute of the storm's motion has the m
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The answer is B. Acceleration.
6 0
3 years ago
Calculate the vapor pressure for a mist of spherical water droplets of radius 3.70×10−8m surrounded by water vapor at 298 K. The
poizon [28]

Answer:

The vapor pressure for a mist is P= 25.92\ Torr

Explanation:

From the question we are given that

        The radius is  r = 3.70 *10^{-8} m

        The temperature is T  = 298K

       The vapor pressure of water P_o = 25.2\ Torr

      The density of water is  \rho = 998 kg.m^{-3}

      The surface tension of water is \sigma  = 71.99 m N \cdot m^{-1}

Generally the equation of that is mathematically represented as

                            ln (\frac{P}{P_0} )  = \frac{2 \sigma M}{r \rho RT}

 Where  P is the vapor pressure for mist

               R is the  ideal gas constant = 8.31

      making P the subject in the formula

    P = e^ {\frac{2 \sigma M}{r \rho RT}} * P_0

        = e^{\frac{2 *(0.07199)(0.018)}{(3.70*10^{-8})(998)(8.31)(298)} } * 25.2

        P= 25.92Torr

               

4 0
3 years ago
Help please !!<br>I need help with this!
belka [17]
Hello Again! I think the Answer might be 220 m! ( 1/2) ( 21 m/s + 0 m/s) (21 s) = 220 m
6 0
3 years ago
Determine the magnitude of the resultant force acting on a 5 −kg particle at the instant t=2 s, if the particle is moving along
Alex787 [66]

Answer:

F = 296.7N

Explanation:

The x and y component of the position vector are given by:

x(t) =rcos\phi, y(t) = rsin\phi\\ \frac{dx}{dt} =\frac{dr}{dt} cos\phi - rsin\phi\frac{d\phi}{dt} , \frac{dy}{dt}=\frac{dr}{dt}sin\phi + rcos\phi\frac{d\phi}{dt}\\ \frac{d^2x}{dt^2} = \frac{d^2r}{dt^2} cos\phi - \frac{dr}{dt} sin\phi\frac{d\phi}{dt} -\frac{dr}{dt} sin\phi\frac{d\phi}{dt} - r(cos\phi\frac{d^2\phi}{dt^2} + sin\phi\frac{d^2\phi}{dt^2})

\frac{d^2y}{dt^2} = \frac{d^2r}{dt^2} sin\phi + \frac{dr}{dt} cos\phi\frac{d\phi}{dt}+\frac{dr}{dt}cos\phi\frac{d\phi}{dt}+r(-sin\phi\frac{d^2\phi}{dt^2} +cos\phi\frac{d^2\phi}{dt^2})

At t = 2s:

\phi(2) = 1.5t^2-6t= -6\\ \frac{d\phi}{dt}(2) = 3t-6=0\\ \frac{d^2\phi}{dt^2}=3\\r(2)=2t+10=14\\ \frac{dr}{dt}=2\\\frac{d^2r}{dt^2} = 0

Plugging in:

\frac{d^2x}{dt^2}=-42(cos(-6) + sin(-6))=-52\frac{m}{s^2} \\\frac{d^2y}{dt^2} = 42(cos(-6)-sin(-6))=28.6\frac{m}{s^2}

The resulting force F is:

F = m\sqrt{(\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2}=296.7N

5 0
3 years ago
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