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4 years ago

A drug like cocaine causes more dopamine to be produced in the brain.

Physics

1 answer:

Andrej [43]4 years ago

8 0

I think it’s true because drugs ruin your system :/

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In which state of matter do the particles have the most energy? (2 points)

jeyben [28]

Gas particles have the most energy.

7 0

3 years ago

Read 2 more answers

Consider a hot air balloon rising vertically from a launch site located on the ground. A person is initially standing 10 m from

abruzzese [7]

Answer:

The distance between the person and the balloon after 2 seconds the person starts walking is changing on 4.47 m/s

Explanation:

The relative position between the balloon and the person is found using Galileo's relativity principle:

$\overrightarrow{r_{b/p}}=\overrightarrow{r_{b}}+\overrightarrow{r_{p}}$ (1)

with Rb the position of the balloon and Rp the position of the person respect with the origin (See Figure 1). Because we don’t have those positions but we know the constant velocities, we can relate the positions (R) with the velocities (v) with the kinematic equation:

$\overrightarrow{R}=\overrightarrow{v}*t$ (2)

So equation (1) is:

$\overrightarrow{R_{b/p}}=\overrightarrow{v_{b}*t}+\overrightarrow{R_{p}}=(v_{b}*t)\,\hat{j}-R_{p\,}\hat{i}$ (3)

with $\hat{j}$ and $\hat{i}$ the unitary vectors on y and x direction respectively.

We see from our initial condition (See figure 2) that:

$R_{p}=(10-v_{p}*t)$ (4)

So if we put this on (3) and divide by time we have:

$\frac{\overrightarrow{R_{b/p}}}{t}=\overrightarrow{v_{b/p}}=\frac{(v_{b}*\cancel{t})}{\cancel{t}}\,\hat{j}-\frac{(10-v_{p}*t)}{t}\hat{i}$ (5)

But we are interested in how fast a distance is changing, and that is a speed so:

$v_{b/p}=\sqrt{v_{b}^{2}+\left(\frac{(10-(v_{p}*t))}{t}\right)^{2}}=\sqrt{4^{2}+\left(\frac{(10-(3*2))}{2}\right)^{2}}\simeq4.47\,\frac{m}{s}$

3 0

3 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it st

andreev551 [17]

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at $70.3^{\circ}$ below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

$y=u_y t+\frac{1}{2}at^2$

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

$u_y=u sin \theta$ is the initial vertical velocity, with u being the initial velocity (unknown) and $\theta=-27.0^{\circ}$ the angle of projection

t is the time of the fall

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Along the x-direction, the equation of motion is instead

$x=(u cos \theta)t$

where $ucos \theta$ is the horizontal component of the velocity. Rewriting this equation as

$t=\frac{x}{ucos \theta}$

And substituting into the previous equation, we get

$y=xtan \theta + \frac{1}{2}gt^2$

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

$t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s$

We can now find the initial speed, u, by using the equation of motion along the x-direction

$x=u cos \theta t$

where we know that:

x = 59.0 m is the horizontal range

$\theta=-23^{\circ}$ is the angle of projection

$t=3.78 s$ is the time of flight

Solving for u, we find the initial speed:

$u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s$

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

$v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s$

The vertical velocity instead changes according to the equation

$v_y = u sin \theta + gt$

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

$v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s$

Where the negative sign means it is downward.

Therefore, the speed at impact is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s$

while the direction is given by

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}$

So, $70.3^{\circ}$ below the horizontal.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

3 years ago

Consider a helium balloon on a string tied to the seat of your stationary car. The windows are closed, so there is no air motion

timama [110]

Answer:

The balloon will move forward.

The density of the air will be greater at the back of the balloon; similar

to the density of air being greater at lower altitudes due to gravitational

attraction because of the weight of the air in an air column.

A block of wood in water rises because of the difference in pressures

on the top and bottom of the block.

6 0

3 years ago

What kind of energy does a light bolt have ?

adell [148]

It has thermal energy

7 0

3 years ago