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3 years ago

Which injury stimulates muscle growth

Physics

2 answers:

krok68 [10]3 years ago

8 0

Microtrauma stimulates muscle growth.

weqwewe [10]3 years ago

7 0

Microtrauma is the answer

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Question 4 of 10

kiruha [24]

B hey what do u know i took that test to

5 0

2 years ago

Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

BlackZzzverrR [31]

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0

3 years ago

A toroid has a square cross section with the length of an edge equal to the radius of the inner surface. The ratio of the magnit

Tasya [4]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

If the distance between the first order maximum and the tenth order maximum of a double-slit pattern is 18 mm and the slits are

Setler79 [48]

Answer:

Wavelength of light is 600 nm

Explanation:

Given

Distance between the first order maximum and the tenth order maximum of a double-slit pattern = 18 mm

Separation between the slits = 0.15 mm

Distance of screen from the slits = 50 cm

Wavelength

$= \frac{18*10^{-3} * 0.15 *10^{-3}}{0.50*9} \\= 6 *10^{-7}\\= 600$ nm

4 0

2 years ago

How much heat is needed to raise the temperature of 8g of water by 15oC?

Murrr4er [49]

<h2>Answer: 502.08 J</h2>

Explanation:

The heat (thermal energy) needed in to raise the temperature in a process can be found using the following equation:

$Q=m.C.\Delta T$ (1)

Where:

$Q$ is the heat

$m=8 g$ is the mass of the element (<u>water</u> in this case)

$C$ is the specific heat capacity of the material. In the case of water is $C=4.184\frac{J}{g\°C}$

$\Delta T=15\°C$ is the variation in temperature <u>(which is increased in this case)</u>

Knowing this, let's rewrite (1) with these values:

$Q=(8 g)(4.184\frac{J}{g\°C})(15\°C)$ (2)

Finally:

$Q=502.08 J$

5 0

3 years ago