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vova2212 [387]
3 years ago
11

During which process do cells use oxygen to release stored energy?

Physics
1 answer:
Grace [21]3 years ago
6 0
I believe it is respiration
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A car having an initial velocity of 12 meters per second speed up uniformly to 22 meters per second. This change is made in 4 se
Sonbull [250]

(22-12)/4

10/4

2.5 meters per second sqaured

7 0
3 years ago
A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=
VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

\Delta x = d-0=d

and the amount of energy required is

U=\frac{1}{2}k d^2

In the second case, the spring is compressed from x=0 to x=-d, so

\Delta x = -d -0 = -d

and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

7 0
3 years ago
When does the radioactive decay of a radioisotope stop? Give one example.
Romashka-Z-Leto [24]

Answer:

When does the radioactive decay of a radioisotope stop? Give one example. An unstable isotope continues the decay process until it reaches a stable form. One example is the decay of carbon-14 to nitrogen-14.

Explanation:

5 0
3 years ago
Read 2 more answers
The elements with the largest atomic radii are found in the
Ulleksa [173]
Elements with the largest atomic radius are found in the lower left hand of the periodic table.
7 0
3 years ago
A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant
Juliette [100K]
<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force F applied to it, <u>as long as the spring is not permanently deformed</u>:

F=k (x-x_{o})    (1)

Where:

k is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

x_{o} is the length of the spring without applying force.

x is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth W of the block:

W=m.g   (2)

Where m=250g=0.25kg is the mass of the block and g=9.8\frac{m}{s^{2}}  is the gravity acceleration.

W=(0.25kg)(9.8\frac{m}{s^{2}})   (3)

W=2.45N   (4)

Knowing the force applied W and x=18cm=0.18m and x_{o}=0, we can substitute the values in equation (1) and find k:

W=k (x-x_{o})    (5)

2.45N=k (0.18m-0m)    (6)

<u>Finally:</u>

k=13.61\frac{N}{m}  

4 0
3 years ago
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