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3 years ago

What is the conclusion of the human impact? Give me at least 3 sentences.

1 answer:

5 0

Human impact on the environment or anthropogenic impact on the environment includes impacts on biophysical environments, biodiversity, and other resources. The term anthropogenic designates an effect or object resulting from human activity. humans poluute,make things that damage the ozone layer. Deforestation has become a huge concern for a wile, we over populate the earth and we change the Environmental behavior because the floods and dams we create.

so:
Humans have a gigantic impact on the world for several reasons, such as; Pollution, deforestation,dams and over population. We deforest the rainforest and create a desert, or desertification , we overpopulate the world, causing a wide spread of pollution;oils,smoke,gases etc.,build dams and change the course of nature, and we alter the ozone layer,creating global warming. Humans have a bigger impact on the world than you might think.

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The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between t

Radda [10]

Answer:

a) $F = 6816.5680 N$

b) $F' = 0 N$

c) $F''=28195.5N$

Explanation:

Magnitude of charges $M=1.6\muC$

Distance $d=2.6$

Generally the equation for Net Force is mathematically given by

For First Drawing

$F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}$

$F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}$

$F = 6816.5680 N$

For second Drawing

$F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}$

$F' = 0 N$

For Third Drawing

$F'' =\frac{ k q^2}{d^2} * \sqrt (2)$

$F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}$

$F''=28195.5N$

4 0

3 years ago

PLEASE ASAP HELP, ILL GIVE WHATEVER POINTS I CAN BUT PLEASE HELP ASAP ASAP

dalvyx [7]

(1) The ball is in the air for 1.4 seconds.

(2) The horizontal velocity of the ball as it rolls off the table is 6.32 m/s.

(3) The vertical velocity of the ball right before it hits the ground is 13.72 m/s.

(4) The horizontal velocity of the ball right before it hits the ground is 6.32 m/s.

(5) The initial vertical velocity as soon as the ball comes of the cliff is 13.72 m/s.

<h3>What is the time of motion of the ball?</h3>

The time of motion of the ball is calculated by applying the following equation.

t = √(2h/g)

where;

h is the height of the cliff
g is acceleration due to gravity

t = √(2h/g)

t = √(2 x 9.63 / 9.8)

t = 1.4 seconds

The horizontal velocity of the ball is calculated as follows;

v = d/t

where;

d is the horizontal distance travelled by the ball = 8.85 m

v = 8.85 m / 1.4 s

v = 6.32 m/s

The vertical velocity of the ball before it hits the ground is calculated as;

vf = vi + gt

vf = 0 + 9.8 x 1.4

vf = 13.72 m/s

The horizontal velocity of the ball right before it hits the ground is calculated as;

the initial velocity of a projectile = final horizontal velocity

vxf = vxi = 6.32 m/s

The initial vertical velocity as soon as the ball comes off the cliff = final vertical velocity = 13.72 m/s

Learn more about horizontal velocity here: brainly.com/question/24949996

#SPJ1

5 0

1 year ago

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

Ymorist [56]

Answer:440.03 N

Explanation:

Given

horizontal component of acceleration $(a_x)=800 m/s^2$

vertical component of acceleration $(a_y)=880 m/s^2$

mass of ball =0.37 kg

Force in horizontal direction $=m\times a_x=0.37\times 800=296 N$

Force in vertical direction $=m\times a_y=0.37\times 880=325.6 N$

Therefore net force is

$F=F_x+F_y$

$|F|=\sqrt{296^2+325.6^2}$

|F|=440.03 N

5 0

3 years ago

When can a truck have less momentum than a bike? A. only when the truck is not moving B. only when it's moving faster than the b

3241004551 [841]

Answer:

Explanation:

It is the answer I think let me know

4 0

3 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[Assume that this experiment takes place in deep space so that the effect of gravity is negligible.]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago