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Mashcka [7]
3 years ago
14

A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a

cceleration of 4m/sec
Physics
1 answer:
Slav-nsk [51]3 years ago
6 0

For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.

For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

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Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

d)  Displacement = 22 m

e)  Average speed = 11 m/s

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a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

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