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-BARSIC- [3]
3 years ago
9

In an energy pyramid, the total energy available for consumption from one trophic level to the next one it is

Physics
1 answer:
Flura [38]3 years ago
5 0

Answer: Conserved

Explanation:

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Demagnetization processes include heating past the Curie point, applying a strong magnetic field, applying alternating current, or hammering the metal.

Explanation:

7 0
3 years ago
Which car is experiencing negative acceleration?
IgorLugansk [536]

Answer:B

Explanation:

4 0
2 years ago
Read 2 more answers
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
HELPP ILL GIVE BRIANLISG
STatiana [176]

Answer:

31. Respiratory System

32. Excretory System

33. Nervous System

34. Reproductive system

35. Skeleton System

Explanation:

Hope it helps you.^_^

6 0
3 years ago
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An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached
Kruka [31]
The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring. 

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be 
<span>F = kx 
270 N/m x 0.38 m = 102.6 N 
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force. 



8 0
3 years ago
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