First to answer will be the brainliest i need the answer ASAP

Find the right answer please

Ghella [55]

Answer:

3.78

Explanation:

bc ik it is :3

5 0

3 years ago

Where is potential energy in food stored

AlexFokin [52]

The potential energy is stored in the chemical bonds of the food. When those bonds break up during the metabolic processes, the energy is released. After that, that energy is stored in the Adenosine Triphosphate bonds aka ATP. The simplest way to think is to think of food as the tightly bound atoms. When the chemical bonds between those atoms break, the stored energy in that food is released.

6 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

3 years ago

As the distance between two charged objects increases, the strength of the electrical force between the objects

GuDViN [60]

Answer:

I believe the answer is It increases

4 0

3 years ago

An hydrogen molecule consists of two hydrogen atoms whose total mass is 3.3×10−27 kg and whose moment of inertia about an axis p

dlinn [17]

Answer:

$6.9631\times 10^{-11}\ m$

Explanation:

I = Moment of inertia = $4\times 10^{-48}\ kg m^2$

m = Mass of two atoms = 2m = $3.3\times 10^{-27}\ kg$

r = distance between axis and rotation mass

Moment of inertia of the system is given by

$I=mr^2\\\Rightarrow I=2mr^2\\\Rightarrow 4\times 10^{-48}=3.3\times 10^{-27}\times r^2\\\Rightarrow r=\sqrt{\frac{4\times 10^{-48}}{3.3\times 10^{-27}}}\\\Rightarrow r=3.48155\times 10^{-11}\ m$

The distance between the atoms will be two times the distance between axis and rotation mass.

$d=2r\\\Rightarrow d=2\times 3.48155\times 10^{-11}\\\Rightarrow d=6.9631\times 10^{-11}\ m$

Therefore, the distance between the two atoms is $6.9631\times 10^{-11}\ m$

3 0

3 years ago