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3 years ago

1. Which of the following statement is correct?

Physics

2 answers:

ElenaW [278]3 years ago

8 0

<h3>1. Which of the following statement is correct?</h3>

A. You should panic when earthquake hits. ❌

B. You should run while the earthquake is going on. ❌

C. You should not follow what the proper authorities will tell you. ❌

$\sf\purple{D.\:You \:should \:prepare\: an \:emergency \:plan\: to \:cope\: with\: the\: disaster. }$ ✅

........................................................................

<h3>2. What should you do during an earthquake?</h3>

A. You should run out of the house. ❌

$\sf\pink{B.\:You\: should \:duck,\: cover, \:and \:hold.}$ ✅

C. You should panic and stay only inside the house. ❌

D. You should lock your door so that nobody can enter. ❌

........................................................................

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

Tanzania [10]3 years ago

3 0

Answer:

D. You should prepare an emergency plan to cope with the disaster.

B. You should duck, cover, and hold.

Explanation:

Because Panic and stress are the main reasons that make the situation worse during natural disasters. Disasters are aggravated when people add chaos to the equation. It's important not to panic during an earthquake. You must maintain a rational state of mind so that you don't do something regretful out of fear.

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How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?

alekssr [168]

Answer:

$1.2646\times10^{25}\ atoms$

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is $Mo(Cr_2O_7)_3$

-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate

-We apply Avogadro's constant to find the number of atoms of oxygen:

$Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms$

Hence, there are $1.2646\times10^{25} \ atoms$

3 0

3 years ago

When observing an object, you determine it is moving at a constant velocity and in a straight line. Which of these BEST describe

Marina86 [1]

The answer would be C. Hope this helps XD

8 0

4 years ago

Read 2 more answers

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

A catapult with a spring constant of 10,000 N/m is used to launch a target from the deck of a ship. The spring is compressed a d

Mnenie [13.5K]

Answer:

Explanation:

Given;

spring constant of the catapult, k = 10,000 N/m

compression of the spring, x = 0.5 m

mass of the launched object, m = 1.56 kg

Apply the principle of conservation of energy;

Elastic potential energy of the catapult = kinetic energy of the target launched.

¹/₂kx² = ¹/₂mv²

where;

v is the target's velocity as it leaves the catapult

kx² = mv²

v² = kx² / m

v² = (10000 x 0.5²) / (1.56)

v² = 1602.56

v = √1602.56

v = 40.03 m/s

v ≅ 40 m/s

Therefore, the target's velocity as it leaves the spring is 40 m/s

6 0

2 years ago

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

4 years ago