All categories

4 years ago

A space shuttle orbits Earth at a speed of 21,000 km/hr. How far does it go in 3.5 hrs?

1 answer:

5 0

Since it travels at 21,000 kilometers per hour, you'd just multiply that with 3.5 to get 73,500. So your answer is 73,500.

You might be interested in

If the frequency of a given wave increases,what happens to the wavelength?

Greeley [361]

It shortens so that the tips reach faster

8 0

3 years ago

Read 2 more answers

Your new pet kitten is trapped in a ten foot deep hole. You need a contraption to safely rescue your poor animal. How can I solv

Amanda [17]

You could use a ladder or a rope

5 0

3 years ago

Distinguish the steps in the process of stopping a tractor, keeping them in proper order.

weeeeeb [17]

Ahhh, there aren’t option to work with :v
Try writing them down here thnx

7 0

3 years ago

Read 2 more answers

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

Zach runs a 100-meter race. 7 seconds after the race started Zach is 35 meters from the starting line and reaches his max speed;

Angelina_Jolie [31]

Answer:MaxSpeed=8 m/s. Additional distance during z seconds after top speed is 8*z meters. Now the distance at 7+z seconds after the race starts is D=( 35+8*z) meters. For x ≥ 7 the time of the race, then D=(35+(x-7)*8) meters is the distance traveled during that time.

Explanation:

First we are interested in calculating the top speed, that is constant. We know at 35 meters he reaches the top speed and this happens 7 seconds after the race started. Also he is at 75 meters from the starting line 12 seconds after the beginning, then we can use the definition of speed, in this case is constant, to get its value: Speed=change in distance / change in time

Then Speed= (75-35)meters/(12-7)seconds = 40m/5s=8m/s. That is the topspeed=8 m/s

Now from the 35 meters he runs at constant speed, then if we are told he runs z seconds and we are asked the distance he runs in that time, we know the distance at constant speed is D=V*t,

then D= 8m/s * z seconds=8z meters.

But here since the run is a 100 meters-race we have an upper bound for z, here he has left (100-35)=65 meters to run at 8m/s, then he can maximum run additional 8,125 second or 15,125 seconds in total for the race.

That is D=8*z meters, with z < 8,125 seconds, this is the additional distance he can run after reaching his top speed.

Now the distance at 7+z seconds after the race starts is D=( 35+8*z) meters, the first 35 m he runs until reaching his top speed and the rest of the distance at this speed.

for x the total time of the run at a given point, we are asked the distance he has traveled at that time, then we are also told x ≥ 7, then D=(35+(x-7)*8) meters, we subtract the 7 initial seconds because they are already counted in the initial 35 meters.

7 0

4 years ago