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bixtya [17]
2 years ago
11

Let's start by calculating what acceleration the rocket must produce to launch into earth orbit. In order to attain orbit around

earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes. What average acceleration is required to accomplish this
Physics
1 answer:
blondinia [14]2 years ago
8 0

Answer:

30.56 m/s^2

Explanation:

Given that In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes.

The average acceleration that is required to accomplish this will be

Average acceleration = change in velocity / time

Average acceleration = 7700/ 4.2 × 60

Average acceleration = 7700/252

Average acceleration = 30.56 m/s^2

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A uniform beam XY is 100 cm long and weighs 4.0N.The beam rests on a pivot 60 cm from end X. A load of 8.0 N hangs from the beam
Alex777 [14]

Answer:

<h2>The magnitude of force F is 18N</h2>

Explanation:

The magnitude of the force in the set up can be solved for using the principle of moment. According to the principle, the sum of clockwise moment  is equal to the sum of anticlockwise moments.

Moment = Force * perpendicular distance

Clockwise moments;

The force that acts clockwise is the unknown Force F and 4N force. If the  beam rests on a pivot 60 cm from end X and a Force F acts on the beam 80 cm from end X, the perpendicular distance of the force F from the pivot is 80-60 = 20cm and the perpendicular distance of the 4N force from the pivot is 60-50 = 10cm

Moment of force F about the pivot = F * 20

Moment of 4N force about the pivot = 4*10 = 40Nm

Sum of clockwise moment = 40+20F...(1)

Anticlockwise moment;

The  8N will act anticlockwisely about the pivot.

The distance between the 8N force and the pivot is 60-10 = 50cm

Moment of the 8N force = 8*50

=400Nm...(1)

Equating 1 and 2 we have;

40+20F = 400

20F = 400-40

20F = 360

F = 18N

The magnitude of force F is 18N

6 0
2 years ago
An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with
Olenka [21]

Answer:

A) 0.957 J

Explanation:detailed explanation and calculation is shown in the image below

8 0
2 years ago
A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca
xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

8 0
2 years ago
An airplane flying at constant air speed from Indianapolis to St. Louis (assume they are directly east-west of each other) in ca
irga5000 [103]

Answer:

Explanation:

Suppose the distance between the two cities is D and the velocity in calm weather is V . The total time taken in two way travel is given by

Total distance / velocity

= 2 D / V

Suppose velocity of wind is v . Then in one way the velocity of airplane will become V + v and in the return trip its velocity will be V - v

Total time taken

= \frac{D}{V+v} +\frac{D}{V-v}

= \frac{2DV}{V^2-v^2}

= \frac{2V^2D}{V(V^2-v^2)}

= \frac{2D}{V(1 - \frac{v^2}{V^2}) }

= The denominator contains a factor

1-\frac{v^2}{V^2}

which is less than one so time calculated will be more than

2D / V

Hence in the second case time taken will be more .

7 0
2 years ago
Calculate the average speed and average velocity of a complete round-trip in which the outgoing 250 km is covered at 95 km/h fol
ser-zykov [4K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.

PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

t = \frac{d}{v}

t = \frac{250km}{95km/h}

t = 2.63h

Time taken for the lunch is

t = 1h

The time taken travel a distance of 250km with a speed of 55km/h

t = \frac{d}{v}

t = \frac{250}{55}

t = 4.54h

The total time taken is

t = t_{outgoing}+t_{lunch}}t_{return}

t = 2.63+1+4.54

t = 8.17h

The average speed is the ratio of total distance and total time

v = \frac{250+250}{8.17}

v = 61.15km/h

PART B)

As the displacement is zero the average velocity is zero.

5 0
2 years ago
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