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a toy car travels from A to B at constant speed 30km/hr and without stopping at B return A at constant speed v. if the average s

adelina 88 [10]

Answer:

Speed BA = 18 km/hr.

Explanation:

Given the following data;

Speed AB = 30km/hr

Speed BA = x km/hr

Average speed = 24 km/hr

To find the value of x;

Average speed = (Speed AB + Speed BA)/2

Substituting into the equation, we have

24 = (30 + x)/2

48 = 30 + x

x = 48 - 30

x = 18 km/hr

4 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

4 years ago

How does the electric force between two charged particles change if one

ExtremeBDS [4]

Answer:

brainly.com/question/11848211

^ Similar/same question as yours!

The pressure between two costs is proportional to the product of the charges.

If solely one of the expenses is decreased by way of a element of 3, then the pressure is decreased with the aid of a thing of 3.

If each prices are reduced by a thing of 3, then the pressure is reduced via a issue of 9.

7 0

4 years ago

A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the cloth

nadya68 [22]

Answer:

The tension on the clotheslines is $T = 8.83 \ N$

Explanation:

The diagram illustrating this question is shown on the first uploaded image

From the question we are told that

The distance between the two poles is $d = 12 \ m$

The mass tie to the middle of the clotheslines $m = 1 \ kg$

The length at which the clotheslines sags is $l = 4 \ m$

Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as

$W = mg$

let the angle which the tension on the clotheslines makes with the horizontal be $\theta$ which mathematically evaluated using the SOHCAHTOA as follows

$Tan \theta = \frac{ 4}{6}$

=> $\theta = tan^{-1}[\frac{4}{6} ]$

=> $\theta = 33.70^o$

So the vertical component of this tension is mathematically represented a

$T_y = 2* Tsin \theta$

Now at equilibrium the net horizontal force is zero which implies that

$T_y - mg = 0$

=> $T sin \theta - mg = 0$

substituting values

$T = \frac{m*g}{sin (\theta )}$

substituting values

$T = \frac{1 *9.8}{2 * sin (33.70 )}$

$T = 8.83 \ N$

6 0

3 years ago

A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0 m/s. If the canyon below is 100.0 m deep, how fa

padilas [110]

If the canyon below is 100.0 m deep, the rocket travels from the edge of the cliff 226m in the horizontal direction.

When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its urface area. Increasing the surface area of an object decreases its speed.

The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

5 0

3 years ago