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Ksenya-84 [330]
3 years ago
12

Answer please need help

Physics
1 answer:
Basile [38]3 years ago
8 0

the answer is C. It allows citizens to submit anonymous tips to the police.

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What do you measure when you measure an object’s mass?
Effectus [21]

Answer:

how much space it takes up in the world

Explanation:

1) Mass is a measurement of the amount of matter something contains, while Weight is the measurement of the pull of gravity on an object. 2) Mass is measured by using a balance comparing a known amount of matter to an unknown amount of matter. Weight is measured on a scale.

3 0
3 years ago
Hello i need help with this!!​
mr Goodwill [35]

Answer:

Yes

Explanation:

If lamp A burnt out there would still be a wire above it that connects lamp B and C to the power source

3 0
2 years ago
A very long straight current-carrying wire produces a magnetic field of 25 µT at a distance d from the wire. How far will the ma
daser333 [38]

The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.

B ∝ 1/d

B = magnetic field strength, d = distance from wire

Calculate the scaling factor for d required to change B from 25μT to 2.8μT:

2.8μT/25μT = 1/k

k = 8.9

You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT

6 0
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
Plz help me ASAP How does sound travel from its source to your ear??? 1. By changes in air pressure 2. By vibrations in wires or
Sholpan [36]

Answer:

2. By vibrations in wires or strings.

8 0
4 years ago
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