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serious [3.7K]
3 years ago
9

A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 29.6 m/s. At the same time, it h

as a horizontal velocity of 54.8 m/s. At what speed does the vehicle move along its descent path?
Physics
1 answer:
inessss [21]3 years ago
6 0

If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is

   √(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2  =  62.28 m/s .

That's 139 mph !  Wow !

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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
A runner begins from rest at the starting line and travles for 6.5 seconds, a runner reaches a speed of 13.4 m/s what is the run
Butoxors [25]

The acceleration of the runner in the given time is 2.06m/s².

Given the data in the question;

Since the runner begins from rest,

  • Initial velocity; u = 0
  • Final velocity; v = 13.4m/s
  • Time elapsed; t = 6.5s

Acceleration of the runner; a = \ ?

<h3>Velocity and Acceleration</h3>

Velocity is the speed at which an object moves in a particular direction.

Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion

v = u + at

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

To determine the acceleration of the runner, we substitute our given values into the equation above.

v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2

Therefore, the acceleration of the runner in the given time is 2.06m/s².

Learn more about Equations of Motion: brainly.com/question/18486505

3 0
3 years ago
Please help me with this question,giving the first correct answer brainliest
Bess [88]

Answer:

Your answer is A

Explanation:

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Answer:

It's either B or D, I'm not positive which it is

Explanation:

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Which example gives all the necessary information to describe motion scientifically
VikaD [51]
Correct option is=A because 25mph means=25miles per hour
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