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3 years ago

Need help girliess.

Physics

1 answer:

steposvetlana [31]3 years ago

8 0

If I remember correctly, it is the 3rd answer choice.

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Give me three examples of fossil fuel-based energy sources pls

svet-max [94.6K]

Gasoline, kerosene, and coal.

4 0

3 years ago

Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show

ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = $\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}$

We know that the inductance of the coil is given by the formula:

$L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH$

4 0

3 years ago

1. Why didn't Sucheng Chan go to school until she was eight

Crazy boy [7]

Answer:

Because china was in war, and her parents didn't have enough money for to go to school.

3 0

3 years ago

3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a

Nonamiya [84]

Answer:

Action - Pulling up the train.

Reaction - Friction on the locomotive

Explanation:

Locomotive is pulling the train upwards ,

Which is the action force applied by the locomotive,

As a reaction locomotive will be pulled by the train which is the reaction of pulling

Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.

For train action is pulling up by locomotive and reaction will be friction acting on it downwards.

6 0

3 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0

3 years ago