True or false a self contained living thing is called adaptation

A bat hasa mads of 2kg at the velocity of 45 m/s what is the kinectic energy could he give to a ball

il63 [147K]

Answer:

the kinetic energy the bat can give to a ball is 2,025 J.

Explanation:

Given;

mass of the bat, m = 2kg

velocity of the bat, v = 45 m/s

The kinetic energy the bat can give to a ball is calculated as;

$K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 2 \ \times \ 45^2\\\\K.E = 2,025 \ J$

Therefore, the kinetic energy the bat can give to a ball is 2,025 J.

8 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

An object that has negative acceleration is definitely doing what?

krek1111 [17]

If an object is consist of a negative acceleration or composed of it, it is likely engaging of having to accelerate in a certain direction in which the direction is heading towards the opposite of the direction that is stated and that is likely to be a positive direction.

3 0

4 years ago

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts

wolverine [178]

Answer:

0.304 m/s2

Explanation:

If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.

We can use Newton's 2nd law to calculate the net acceleration of the system

$a = F/m = 7 / 23 = 0.304 m/s^2$

5 0

3 years ago

A runner makes one lap around a 200 m track in a time of 25.0 s. What was the runner's average speed?

creativ13 [48]

The answer is 8 meters per second 8m/s because 200 divided by 25 is 8

3 0

4 years ago

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