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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

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

Learn more about current and voltage:

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#LearnwithBrainly

4 0

3 years ago

A jet with mass m = 90000.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.

leva [86]

This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)

6 0

3 years ago

On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question

romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(c) 45.4545 kg on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

(c) We have given that weight of the person on the earth = 100 lbs

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs $\frac{100}{0.37}=270.27kg$

8 0

3 years ago

N LC circuit has an oscillation frequency of 105 Hz. If C = 0.1 F , then L must be about:

Umnica [9.8K]

Answer:

L = 22.97 H

Explanation:

Given that,

Capacitance, $C=0.1\ \mu F=0.1\times 10^{-6}\ F$

Oscillation frequency, f = 0.5 Hz

The frequency of an AC circuit is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

Where

L is impedance

$f^2=\dfrac{1}{4\pi ^2LC}\\\\L=\dfrac{1}{4\pi ^2 f^2 C}\\\\\text{Putting all the values}\\\\L=\dfrac{1}{4\pi^2 \times (105)^2\times 0.1\times 10^{-6}}\\\\L=22.97\ H$

So, the impedance of LC circuit 22.97 H.

7 0

2 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

3 years ago