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Andrej [43]
2 years ago
12

If the resistance through the fat is 3 times that through the muscle, how much of the total current goes through the fat in term

s of the current through the muscle?
A) 3 times the current through the muscle.
B) The same current as through the muscle.
C) No current passes through the fat.
D) 1/3 times the current through the muscle.
Physics
1 answer:
lapo4ka [179]2 years ago
8 0

Answer:

it A i think

hope that helps if not i can change it

Explanation:

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What effect does observing a substance's physical properties have on the substance?
Free_Kalibri [48]

If you're careful, you ought to be able to observe ANY of these properties
without any effect on the substance:

Absorption, albedo, angular momentum, area, color, concentration,
density, elasticity, electric charge, electrical conductivity, flow rate,
electrical impedance, electric potential, fluidity, length, location, mass,
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permeability, permittivity, plasticity, pressure, radiance, solubility, spin,
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6 0
3 years ago
A cannon fires a cannonball at a 35.0° angle at 62.0 m/s on level ground. (a) What is the maximum height of the cannonball? (b)
ycow [4]

Answer:

a) The maximum height of the cannonball is 64.5 m.

b) The cannonball´s speed at maximum height is 50.8 m/s.

Explanation:

The position and velocity vectors of the cannonball can be calculated using the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity vector at time t.

a) At the maximum height, the vertical component of the velocity vector is 0 (please, see the attached figure and notice that at the maximum height the velocity vector is horizontal).

Knowing this, we can calculate the time at which the cannonball is at its maximum height:

vy = v0 · sin α + g · t

0 m/s = 62.0 m/s · sin 35.0° - 9.81 m/s² · t

- 62.0 m/s · sin 35.0° / -9.81 m/s² = t

t = 3.63 s

Now, we can calculate the y-component of the vector r1 in the figure (r1y):

y = y0 + v0 · t · sin α + 1/2 · g · t²

The cannon is at the same level that the origin of the frame of reference (the ground) so that y0 = 0.

y = 0 m + 62.0 m/s · 3.63 s · sin 35.0° - 1/2 · 9.81 m/s² · (3.63 s)²

y = 64.5 m

The maximum height of the cannonball is 64.5 m

b) To calculate the speed at the maximum height, we can use the equation of the velocity vector:

v = (v0 · cos α, v0 · sin α + g · t)

We already know that the y-component is 0. Then, let´s calculate the x-component of the velocity:

vx = v0 · cos α

vx = 62.0 m/s · cos 35.0°

vx = 50.8 m/s

The vector velocity at maximum height will be:

v = (50.8 m/s, 0)

The speed is the magnitude of the velocity vector:

|v| = \sqrt{(50.8 m/s)^{2} + (0 m/s)^{2}} = 50.8 m/s

The cannonball´s speed at maximum height is 50.8 m/s.

3 0
3 years ago
In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.
garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...

Find the equivalent resistance of the right branch of the circuit:

R_{eq}  = R_{2} +R_{3} \\R_{eq} = 20.7 + 25.8 = 46.5 ohms

\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411

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A bullet travels at 850 m/s. How long will it take a bullet to go 1 km?
Stells [14]
The answer to your question is 1.176 seconds

7 0
3 years ago
Students are experimenting with circuits in their physics class and they build the two working circuits pictured below. The batt
bija089 [108]

The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs.  In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases.  In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.

<h3>What is power of the circuit?</h3>

The power of the bulb or any resistor is equal to the product of voltage and  current flowing through it.

P = VI

Circuit A has bulbs in series while the circuit B has bulbs in parallel.

When bulb 3 added to circuit A,  the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.

The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.

Thus, the last option is correct.

Learn more about power.

brainly.com/question/2933971

#SPJ1

4 0
2 years ago
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