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2 years ago

Compare the benefits of wildfires to grasslands, northern forests, and deciduous forests.

2 answers:

6 0

Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.

zhannawk [14.2K]2 years ago

6 0

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Drag the tiles to the correct boxes to complete the pairs. Match the sentences with the steps of the scientific method

Assoli18 [71]

Solution:

Make an Observation - An indoor plant in a dark room withers faster than the same plant in a room with ample sunlight.

Ask a question- Why do certain indoor plants die faster based on where they are placed in the house?

State a hypothesis- Sunlight is probably essential for plants to grow and live.

Run an experiment- Get two potted plants. Cover one with black paper. Place both plants outside in sunlight. See what happens to each plant after one week.

Analyze the results-The plant in the pot with black paper withered. The other plant was healthy.

Communicate the results to others - Plants need sunlight to make food so they can live.

4 0

2 years ago

One nucleus contains 31 protons and 40 neutrons another nucleus contains 31 protons and 41 neutrons what can you conclude about

Pavel [41]

They are isotopes of the same element.

5 0

2 years ago

Read 2 more answers

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w

Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg mass of the box

d= 5.4m length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.

W: Weight of the box : In vertical direction

N : Normal force : perpendicular to the direction the roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight of the box

W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86 = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34 m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:

d:displacement = 5.4 m

v₀: initial speed

vf: final speed = 0

a : acceleration of the box = -1.34 m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

$v_{o} =\sqrt{14.472}$

v₀ = 3.8 m/s

7 0

2 years ago

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to fi

JulijaS [17]

Answer:

b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

w₀² = k / m

w² = w₀² - b² / 4m²

b² = (w₀² -w²) 4 m²

Let's find the angular velocities

w₀² = 5 / 0.1

w₀² = 50

w = 2π f

w = 2π 1

w = 6.2832 rad / s

we subtitute

b² = (50 - 6.2832²) 4 0.1²

b = √ 0.42086

b = 0.6487 kg / s

8 0

2 years ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car

Novay_Z [31]

Answer:

$4.5 *10^{3}\ kgm/s^{2}$

Explanation:

Given:

$Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\ meters/second^{2}$

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration we get ,

$Force\ = \ 1.00 * 10^{3} * 4.5\ m/s^{2}$

$Force\ =4.5 * \ 10^{3} \ kgm/s^{2}$

Therefore Net force is : $4.5 *10^{3}\ kgm/s^{2}$

0 0

2 years ago