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vampirchik [111]
3 years ago
11

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

d to change the velocity of a(n) 83000 kg space craft by 71 cm/s, how long should it be fired? Answer in units of s
Physics
1 answer:
Over [174]3 years ago
7 0

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

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Find the area of the part of the plane 5x 3y z = 15 that lies in the first octant.
Lunna [17]

This part of the plane lies above a triangle with boundaries x=0 and y=0 along the coordinate axes, as well as the line

z=0 \implies 5x + 3y = 15 \implies y = \dfrac{15 - 5x}3

When y=0, we have 15-5x=0\implies x=3. So this triangle is the set

T = \left\{(x,y)\in\Bbb R^2 ~:~ 0\le x\le3 \text{ and } 0\le y\le\dfrac{15-5x}3\right\}

Also, when x=0, we have y=\frac{15}3=5. So the triangle has length 3 and width 5, hence area 1/2•3•5 = 15/2.

Let z=f(x,y) = 15 - 5x - 3y. Then the area of the plane over T is

\displaystyle \iint_T dA = \iint_T \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy

We have

\dfrac{\partial f}{\partial x} = -5 \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 25

\dfrac{\partial f}{\partial y} = -3 \implies \left(\dfrac{\partial f}{\partial y}\right)^2 = 9

\implies\displaystyle \iint_T dA = \sqrt{35} \iint_T dx\,dy = \boxed{\frac{15\sqrt{35}}2}

since the integral

\displaystyle \iint_TdA

is exactly the area of T, 15/2.

8 0
2 years ago
HELP PLS and HURRY
tangare [24]

Answer:

C - 50,000 * 77 * 3

Explanation:

At the top of the hill the potential energy is E= mgh= (160 kg)(9.81 m s^-2)(30 m)= 47088

hope it helps ,

<u>help me by marking as brainliest....</u>

5 0
3 years ago
In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us
dybincka [34]

Answer:

1.95m/s

Explanation:

Please view the attached file for the detailed solution.

The following were the conversion factors used in order to express all quatities in SI units:

1 gallon=0.00378541m^3\\1 inch=0.0254m\\1 minute=60s

6 0
3 years ago
Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus
poizon [28]

Answer:

For destructive interference phase difference is

(2n+1)\pi where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of \frac{\pi}{4} would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of  ( \frac{\pi}{4} + 2\pi) because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of \pi or which is same as (2\pi+\pi)

Generalizing to:

a phase difference of (2n+1)\pi where n∈ {W}

{W}= set of whole numbers.

3 0
3 years ago
Using this information...
Pepsi [2]

19.2\:\text{m/s}

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)

\:\:\:\:\:= 19.2\:\text{m/s}

6 0
3 years ago
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