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bixtya [17]
4 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]4 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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Explain the effects of force giving example​
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1) Can change the state of an object(rest to motion/ motion to rest):For example, pushing a heavy stone in order to move it. 2) May change the speed of an object if it is already moving. 4) May bring about a change in the shape of an object. For example, blowing air in balloon.

3 0
3 years ago
Read 2 more answers
A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The
Liula [17]
What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be \frac{F}{2} now:

a=\frac{F}{2m}

So the accelaration will also he halved (it's the original acceleratation divided by 2)


 b) The object's mass is halved?
a=\frac{F}{m/2}=a=\frac{F2}{m}

which is the original acceleration times two!! so it will double


c) The force and the object's mass are both halved?
now we have

a=\frac{F/2}{m/2}=a=\frac{2F}{2m}=a=\frac{F}{m}

so they will cancel each other out and the acceleration will stay the same!











5 0
3 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
4 years ago
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