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3 years ago

Which of the following statements would be classified as an aphorism?

1 answer:

5 0

A bird in the hand is worth two in the bush
or
actions speak louder than words
or
her bark is worse than her bite
i hope this helps these are examples of aphorism

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Currents during lightning strikes can be up to 50000 A (or more!). We can model such a strike as a 47500 A vertical current perp

Rufina [12.5K]

Answer:

57 N

Explanation:

Force on a current carrying conductor in a magnetic field

B = 12 X 10⁻⁴ T

= Bil where B is magnetic field , i is current and l is length of conductor

force required = 12 x10⁻⁴ x 47500 x 1

= 57 N

6 0

3 years ago

True or false it is impossible to determine weather you are moving unless you can touch another object

Lerok [7]

Answer: false

Explanation:

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3 years ago

Read 2 more answers

List two industrial uses for nonmetallic mineral resources

Anni [7]

Answer:

To make useful chemicals and abrasives

Explanation:

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3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

The electric field between two parallel plates connected to a 45-volt battery is 500. volts per meter. The distance between the

kotykmax [81]

Electric field = potential difference
-----------------------------
distance between plates

Distance between plates = 45
----------
500

= 0.09 meters.

8 0

3 years ago