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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 !

You might be interested in
If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
solong [7]
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s

3 0
2 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

8 0
2 years ago
A mechanic uses a jack to lift up a car. He exerts a force of 11,000 N at a distance of 3m from the axis of rotation. How much t
pshichka [43]

Answer:

<h2>The amount of torque put on the car is 33,000Nm</h2>

Explanation:

Formula for calculating torque is expressed as T = rFsin\theta\\ where;

r is the radius of the  of the arm of the jack = 3m

F is the force exerted = 11000

\theta\\ is the angle of rotation = 90°

On substituting;

T = 3*11000sin90^{o} \\T = 3*11000 (sin90^{o} =1)\\T = 33000Nm

6 0
3 years ago
Science is the human endeavor to understand the natural world true or flase
hjlf
True yes TRUE
Science may also be defined as the study of surroundings
4 0
3 years ago
two forces 3N and 4N act on a body in a direction due north and due East respectively calculate their equivalent​
Lostsunrise [7]

Two forces 3N and 4N act on a body in a direction due north From East, the equilibrant's angle is given by \theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}.

<h3>What are equilibrium and resultant force?</h3>

The equilibrium force is the balanced force when the net force acting is zero and is the exact opposite of the consequent force. The resultant force is one single force replaced by numerous forces.

<h3>Briefing:</h3>

3N and 4N are the two forces pulling on a body.

The forces work along the North and the East, which are perpendicular to one another.

The resultant of the forces, which is provided by the equilibrant force,

R  = √(3)²+(4)²

R = 5N

From East, the equilibrant's angle is given by

\theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}

To know more about equilibrium force visit:

brainly.com/question/12582625

#SPJ9

6 0
11 months ago
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