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1 year ago

The image shows positions of the earth and the moon in which region would an astronaut feel the lightest

Physics

2 answers:

DanielleElmas [232]1 year ago

8 0

Answer:

The moon region

Explanation:

trapecia [35]1 year ago

5 0

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

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Bernice draws an oxygen atom. She draws a small circle for the nucleus. Inside of the circle, she draws plus signs for protons a

Lelu [443]

Answer

It should be A and C

Explanation:

because oxygen is number 8 in the periodic table of elements and has a atomic weight of 15.999 you use those numbers to figure out what is true between those.

The 8 for oxygen goes for the number of electrons and proton and to find neutrons u round the 15.999 up which now make it 16 and subtract it by the 8 now you have 8 protons, 8 neutrons, and 8 electrons

5 0

1 year ago

Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa

Katarina [22]

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%

8 0

1 year ago

It takes a slug 20 minutes to travel from the grass to the trash can , a trip of 15 meters. How far could the slug travel in 60

inn [45]

Answer:

45 meters

Explanation:

20 min = 15 meters

So if 20 x 3 = 60

you have to do 3 x 15 !

- which equals to 45 <3

<u>- mark me brainlest pls . </u>

5 0

2 years ago

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0

2 years ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

1 year ago