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ch4aika [34]
3 years ago
6

An electroscope is a device that has small gold foil strips suspended from a metal rod. A student brings a negatively charged ro

d near the top of a neutral electroscope. The student then briefly touches a hand to the top of the electroscope before removing the charged rod. The foil strips of the electroscope now spread apart. The student then brings a different rod near the top of the electroscope. The leaves of the electroscope move closer together (see below). The new rod must have a net charge that is
1) positive.
2) negative.
3) neutral.
4) not enough information.
Physics
1 answer:
Sever21 [200]3 years ago
4 0
I think it is neutral
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Two positive charged particles will ?
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Answer:

make a negative

Explanation:

yep

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3 years ago
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A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
3 years ago
If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

Explanation:

The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

3 0
3 years ago
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Nostrana [21]

Answer: The heart pumps 124.2 billion cm³ of blood in a lifetime

Explanation:

as an adult the pulse rate average must be around 72 beats per minute.

The heart beats about 103,680 times in a day.

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number of heart beat in a year = 365 days x 103,680 = 37,843,200 beats in a year

For every the heart pumps 50cm³ of blood,

Hence,

Amount of blood pump in a year = 50 x 37,843,200 = 1,892,160,000cm³ of blood pumped in a year.

Using the estimated lifespan average an individual is 69 years

So in a life time,

The human heart pumps = 1,892,160,000 x 69 years = 124,200,000,000

If the heart pumps 50cm³ of blood per beat, the heart pumps a total of 130,559,040,000 cm³ (130.6 billion cm³) of blood in a LIFETIME.

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