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3 years ago

During an earthquake, you should do all of the following EXCEPT _________________________.

2 answers:

6 0

The key word is "electric". If the power goes out during an earthquake, items that run on electricity won't help you any. You need battery-operated radios and flashlights, as well as extra batteries.
What are your answer choices?

frosja888 [35]3 years ago

4 0

Run from the area in a zig-zag formation or keep hands visible. if these are wrong then sorry.

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In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves

11111nata11111 [884]

Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m
m = 280 kg 
Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N

1) What is the magnitude of the normal force on the care when it is at the bottom of the circle.

Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N 

2) What is the magnitude of the normal force on the car when it is at the side of the circle. 

Centripetal force = 7119.55 N 

3) What is the magnitude of the normal force on the car when it is at the top of the circle. 

Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N 

4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. 

√(gr) 

√(9.8 x 11.5) = 10.62 m/s

7 0

3 years ago

A.) Determine the work done by Zach on the bull.

tresset_1 [31]

Explanation:

The unit of work done is in Joules

Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

Work done = Force x distance

The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

So to solve this problem;

multiply the force applied by Zack and distance through which the bull was pulled.

8 0

3 years ago

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

$h=ut+\frac{1}{2}at^2$

where

$h=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

here $u=0$ because it dropped from a certain height

$h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

When height is increases to three times of original height

i.e. $h'=3 h$

then time period becomes

$T'=\sqrt{\frac{2\times 3h}{g}}$

$T'=\sqrt{3}T$

5 0

3 years ago

A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0

3 years ago

Cocaine, which is derived from the coca leaf, is grown mostly in Colombia and its neighboring countries while heroin, which is d

nevsk [136]

Answer:

Because of the location, humidity and temperatures.

Explanation:

Coca is grown in humid and very humid subtropical forests, called yungas and

they form the lower floor of the upper Jungle, in the Central Andes, mostly in Peru and Bolivia. The yungas are in contact with the rainforests of the lowlands in Amazonia, where it has been started to expand coca cultivation recently (Dourojeanni, 1988). The optimum altitude is 1000 a 2000 meters (where cocaine content is higher), with optimal annual average precipitation, is 2000 meters mm, but it is grown between 700 and 2000 msnm and with an average annual rainfall of 1000 to 4200 mm.

msnm = meters above sea level

6 0

3 years ago