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3 years ago

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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11) A sled is initially given a shove up a frictionless 35º incline. It reaches a maximum height of 2.5

Andrej [43]

Answer:

7 m/s

Explanation:

To solve this problem you must use the conservation of energy.

$K1 +U1=K2+U2$

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.

The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:

$KE1= \frac{1}{2}mv^{2}}$

$PE2=mgh$

K1=U2

$\frac{1}{2} mv^{2} =mgh$

Solve for v

$v=\sqrt{2gh}$

Input known values and you get 7 m/s.

5 0

2 years ago

A grapefruit falls from a tree and hits the ground 0.76 s later.

Tomtit [17]

Answer:

2.8 m 7.4 m/s

Explanation:

write all the values then use the equations of motion to find the distance and speed. please see attached photo

7 0

3 years ago

How many Joules of potential

s2008m [1.1K]

Answer:

40 j, 80j.

Explanation:

P.E= mgh. G=10 m/s².

For 4m, P.E=1*10*4=40 joules.

For 8m, P.E=1*10*8=80 joules.

4 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago

Formula for deriving initial temperature in linear expansivity

jekas [21]

Answer:

∆L=aL∆T

Explanation:

that's the answer for your Question

5 0

2 years ago