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3 years ago

14

What would result from under-coverage? A. a lack of accurate result B. an incorrect independent variable C. an experiment D. acc

urate result

2 answers:

ohaa [14]3 years ago

4 0

<span>What would result from under-coverage?
the answer is A.)</span><span>a lack of accurate result</span>

kherson [118]3 years ago

3 0

A. a lack of accurate <span>result</span>

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In an experiment involving pendulums, you want to see how changing the mass of the bob affects the period (amount of time) of a

Alborosie

Answer:

The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.

Explanation:

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2 years ago

A guitar string with a linear mass density of 2.0 g/m is stretched between two vertical rods that are 65 cm apart. The string ho

Rudiy27

Answer:

717 Hz

Explanation:

<u>solution:</u>

The wave with three antinodes has m = 3. Thus, f_3 = 3f_1 and so the fundamental frequency of the string is

f_1 =f_3/3

=430 Hz/3

=143 Hz

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3 years ago

Light travels 300 000 000 m/s and one year has approximately 32 000 000 second a light year is the distance light travels in one

Lelu [443]

Explanation:

It is given that,

Speed of light, $v=300 000 000\ m/s=3\times 10^8\ m/s$

Seconds in 1 year, $t=32 000 000=32\times 10^6\ s$

We need to find the distance traveled by light in one year. Speed of an object is given by :

$v=\dfrac{d}{t}$

So,

$d=v\times t\\\\d=3\times 10^8\times 32\times 10^6\\\\d=9.6\times 10^{15}\ m$

Since,

$1\ \text{light year}=9.46\times 10^{15}\ m\\\\1\ m=\dfrac{1}{9.46\times 10^{15}}\ \text{ly}\\\\9.6\times 10^{15}\ m=\dfrac{9.6\times 10^{15}}{9.46\times 10^{15}}\\\\d=1.01\ \text{ly}$

So, the distance covered by light is 1.01 light years.

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3 years ago

The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed o

SOVA2 [1]

Answer:

50.97 m

Explanation:

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$\mu_s$ = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = $\dfrac{72}{3.6}=20\ \text{m/s}$

s = Displacement

Force applied

$F=ma$

Frictional force

$f=\mu_s mg$

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

$ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2$

Since the obect will be decelerating the acceleration will be $-3.924\ \text{m/s}^2$

From the kinematic equations we have

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}$

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

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3 years ago

Energy conversion from potential energy (the energy of position or composition) to kinetic energy (the energy of motion) is illu

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Answer:

When it goes up it gains PE and when it moves down PE is converted into KE

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3 years ago