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1 year ago

If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.

1 answer:

7 0

V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance.

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Describe each class of lever and explain to characteristics of each

Nataly [62]

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example: a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example: a nut-cracker, a garlic press

-- Class III lever

The effort is between the fulcrum and the load.

The Mechanical Advantage is always less than 1 .

I can't think of an example right now.

8 0

1 year ago

Read 2 more answers

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

$0 = m(3.5 sin30) - mv_y$

$v_y = 1.75 m/s$

Now similarly in X direction we will have

$m(4.5) = m(3.5 cos30 ) + mv_x$

$v_x = 1.47 m/s$

now final kinetic energy of both puck after collision is given as

$KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)$

$KE_f = 8.73 m$

initial kinetic energy of both pucks is given as

$KE_i = \frac{1}{2}m(4.5^2) + 0$

$KE_i = 10.125 m$

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0

2 years ago

For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?

Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

$85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}$

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity ( $v_t$ ), in order to obtain the angular velocity of the wheel:

$v_t=w*R\\w=\frac{v_t}{R}$

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

$w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}$

And now, recalling that $2\pi$ radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by $2\pi$ :

rotational frequency = $\frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}$

6 0

2 years ago

Most of the nutrients in the rainforest ecosystem are in the _____.

joja [24]

<span>The answer should be the vegitation. </span>

4 0

1 year ago

Read 2 more answers

A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

1 year ago