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2 years ago

DC versus AC problem. Suppose your DC power supply is set to 15 V and the vertical setting on the oscilloscope is at 5 V/div.

Physics

1 answer:

Lana71 [14]2 years ago

7 0

Answer:

DC = 3 div. AC= 0

Explanation:

When the input is directed to the input circuitry, and the "A input" switch is set to DC, the pure 15 V DC signal will be showed on the screen, so, if the vertical setting is at 5 V /div, the trace will deflect exactly 3 div.

If the switch is on AC, as this setting inserts a capacitor in series (which is located here to block any unwanted DC component superimposed to an AC signal) the DC signal will be blocked, so no trace will be deflected on the screen (after completed the transient period).

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You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

3 0

2 years ago

A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete

erastovalidia [21]

Answer:

407 steps

Explanation:

From the question,

P = mgh/t........... Equation 1

Where P = power, m = mass, g = acceleration due to gravity, h = height, t = time.

Make h the subject of the equation

h = Pt/mg............. Equation 2

Given: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Constant: g = 9.8 m/s²

Substitute into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

number of steps = 6525/16

number of steps = 407 steps

6 0

2 years ago

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago

A particle is in uniform circular motion. Assume a standard rtz coordinate system. If you deconstruct the net force acting on th

Kitty [74]

Answer:

a) One

Explanation:

In a uniform circular motion there must be a force acting to keep it in the circular track. This force can either be centripetal or a centrifugal force.

According to the Newton's first law of motion a particle continues to be in state of rest or in uniform motion until acted upon by an external force.

Here the term uniform motion need to be understood which refers to the uniform velocity of the particle in accordance to the vector laws.

3 0

2 years ago

In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0

2 years ago