2 years ago

Which best esplains how the body maintains homeostasis.

Physics

2 answers:

Lisa [10]2 years ago

8 0

a. all systems work together to stabilize the body

CaHeK987 [17]2 years ago

3 0

All systems work together to stabilise the body!!

You might be interested in

What was the average velocity for the entire trip?

just olya [345]

Answer:

This is the explanation of the ans

6 0

2 years ago

Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a

dybincka [34]

I believe it’s divergent boundary but I might be wrong

5 0

3 years ago

Read 2 more answers

oe finds that the temperature of a substance is 12 degrees Celsius. What does this tell Zoe about the substance? Its internal en

Roman55 [17]

Its internal energy is less than 12 degrees

5 0

3 years ago

Read 2 more answers

In this football play the quarterback keeps the ball and goes straight ahead, behind the blocking of the center and the guards.

oksano4ka [1.4K]

Answer:

Explanation:

3 0

3 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

2 years ago