Ask question

Ask question

All categories

3 years ago

10

If you were creating a presentation that would cover the different types of muscles in the body, which groups of muscle would yo

u need to include?

2 answers:

diamong [38]3 years ago

6 0

It would be best to cover the cardiac, smooth, and skeletal muscles! =)

Flura [38]3 years ago

3 0

Answer:

c

Explanation:

You might be interested in

When rocks experience high temperatures and differential stresses deep in the earth, their grains tend to ________?

pochemuha

Break into small particles. The heat causes the rock to break up and form pebbles or sand. Hope that helped. Have a nice day

4 0

3 years ago

A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

podryga [215]

I believe it is -1.11 m/s^2. I will let you know if its correct

7 0

3 years ago

Read 2 more answers

100 POINTS!! PLEASE DONT ANSWER UNLESS YOU KNOW BOTH ANSWERS PLEASE.

JulijaS [17]

Answer:

1. is B and 2. is A

Explanation:

6 0

3 years ago

Read 2 more answers

Blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up in a row on a frictionless table. All three are pushed forward by

exis [7]

Answer:

This is very hard bit I think 6.3 my, I'm not shure.

5 0

3 years ago

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

3 years ago