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2 years ago

Difference between global warming and climate change

Physics

1 answer:

KIM [24]2 years ago

6 0

Answer:

Global warming refers only to the Earth’s rising surface temperature, while climate change includes warming and the “side effects”

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A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H

IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm $f_{0} = 1210\ Hz$

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

$f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})$

Where, $f_{0}$ = frequency of source

$v_{0}$ = speed of observer

$v_{s}$ = speed of source

$v$ = speed of sound

Put the value into the formula

$f=1210\times\dfrac{343+(-8.25)}{343-0}$

here, source is at rest

$f=1180.8\ Hz$

$f=1181\ Hz$

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0

3 years ago

An atom of arsenic has how many electron-containing orbitals

irina [24]

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

$1s^{2}$

$2s^{2}$

$2p^{6}$

$3s^{2}$

$3p^{6}$

$3d^{10}$

$4s^{2}$

$4p^{3}$

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

8 0

3 years ago

What is the moment of inertia of the object starting from rest if it has a final velocity of 5.9 m/s? Express the moment of iner

Bogdan [553]

Answer:

The moment of inertia is I = 0.126*R^2*M

Explanation:

We can calculate the moment of inertia of an object that starts from rest and has a final velocity using the energy conservation equation, as follows:

Ek1 + Ep1 = Ek2 + Ep2, where

Ek1 = kinetic energy of the object before to roll down

Ep1 = potential energy of the object

Ek2 = kinetic energy when the object comes down

Ep2 = potential energy of the object at the bottom

We have the follow:

Ek1 = 0

Ep1 = M*g*h

Ek2 = ((I*w)/2) + ((M*v^2)/2)

Ep2 = 0

Replacing values:

0 + M*g*h = ((I*w)/2) + ((M*v^2)/2) + 0

where:

M = mass of the object

g = gravitational acceleration

I = moment of the inertia

w = angular velocity = v/R

h = height

M*g*h = ((1/2) * I * (v^2/R^2)) + ((M*v^2)/2)

M*9.8*2 = (I*(5.9^2)/(2*R^2)) + ((5.9^2 * M)/2)

19.6 * M = ((17.4*I)/R^2) + 17.4*M

Clearing I, we have:

I = 0.126*R^2*M

5 0

2 years ago

A man lifts a 120 kg barbell 2 m above the ground . What is the gain in gravitational PE of the barbell?

SpyIntel [72]

Answer:

2,352 Joules

Explanation:

At the ground, the barbell has a classical mechanical energy value of zero. There is no classical kinetic or potential energy for the barbell. The moment the man starts to lift the barbell, he does work on the barbell and transfers kinetic energy to it due to the motion. At its maximum height where the man lifts the barbell to a stop, the kinetic energy is zero because it transformed into gravitational potential energy stored in the gravitational field. Our reference point for potential was defined to be zero at the floor, therefore we can say that the gravitational potential energy at 2 meters is:

$U=mgh=(120kg)(9.8m/s^2)(2m)=2,352J$

7 0

2 years ago

An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act

seropon [69]

Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector $\hat{i}$ .
The positive y-axis = the northern direction, with unit vector $\hat{j}$ .

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
$\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}$

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
$\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})$

The plane's actual velocity is the vector sum of the two velocities. It is
$\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}$

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.

8 0

3 years ago