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3 years ago

Human activities can have an impact on natural disasters.

2 answers:

6 0

I think you wanted to know whether the statement in question is true or false. Assuming this fact, i am answering the question and hope that it helps you. It is a true fact that human activities can have an impact on natural disasters. The increase in global temperature due to emission of greenhouse gases can cause the sea level to rise and create hurricanes.

alukav5142 [94]3 years ago

4 0

By definition, we have to:

The environmental impact are the consequences that some of the human activities produce on the environment.

The concept can be extended to the effects of a catastrophic natural phenomenon.

Human activities, such as fuel combustion and deforestation, have intensified the greenhouse effect, causing global warming.

It has been estimated that if emissions of greenhouse gases continue at the current rate, the temperature of the planet could exceed historical values in 2047, with harmful effects on ecosystems, biodiversity and endanger the livelihood of people on the planet.

Answer:

True.

Human activities can have an impact on natural disasters.

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A _____ is a model of an atom in which each dot represents a valence electron

Rainbow [258]

Answer:

electron dot diagram is the answer to your question.

6 0

3 years ago

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑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)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago

A helium balloon has a pressure of 40 psi at 20°C. What will the pressure be at 40°C? Assume constant volume and mass.

Likurg_2 [28]

Answer:

P V = N R T ideal gas equation

P2 / P1 = T2 / T1 where the other variables are constant

P2 = (T2 / T1) * P1 = (313 / 293) * 40 psi = 42.7 psi

5 0

2 years ago

You have been asked to make a roller coaster more exciting. The owners want the speed at the bottom of the first hill doubled. H

qaws [65]

Answer:

The height will be 4 times.

Explanation:

Given that,

The speed at the bottom of the hill doubled.

We need to calculate the height

Using conservation of energy

$K.E_{t}+P.E_{t}=K.E_{b}+P.E_{b}$

$K.E_{b}=P.E_{t}$

$\dfrac{1}{2}mv^2=mgh$

$\dfrac{1}{2}m(4v^2)=mgh$

Therefore,

$P.E = mg(4h)$

Here, m and g are constant

Hence, The height will be 4 times.

4 0

3 years ago

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s

skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0

3 years ago