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Irina-Kira [14]
3 years ago
6

28. A student notices that when a weight is hung on a spring, the spring stretches. She decides to conduct an experiment to dete

rmine the relationship between the amount of weight
placed on a spring and the distance the spring stretches. She has five different weights: 25 9,50 9.75 9. 100 9, and 125 g. She selects a weight, hangs it on the spring, and
measures how far the spring stretches
What is the independent variable in this experiment?
length of unst etched spring
distance the spring stretches
weight hung from the spring
O temperature of the spring
Physics
1 answer:
azamat3 years ago
3 0

Answer:

C). weight hung from the spring

Explanation:

An Independent variable is characterized as the variable that is controlled or manipulated by the experimenter in order to observe its change on the dependent variable.

In the given experiment, 'weight hung from the spring' is the independent variable as it is manipulated by the researcher in order to witness its impact on the dependent variable(how the spring stretches). It is the cause that causes an effect on the dependent variable. Any change in the former directly affects the latter. Thus, <u>option C</u> is the correct answer.

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Use scientific (exponential) notation to express the following quantities in terms of the SI base units in
Anna007 [38]

Answer:

When a number is written in scientific notation (representing the number using powers of base ten) it is expressed so that it contains a digit in the place of the units and all other digits after the decimal point, multiplied by the respective exponent.  For example, the number 4.232(10)^{3}.

On the other hand, it is known the units in the SI for mass, length, time and temperature are kilogram (kg), meter (m), second (s) and Kelvin (K), respectively. In addition, thera are prefixes of the International System (SI) that indicate a specific factor of 10.

For example:

-Giga (G) is a prefix that indicates a factor of 10^{6}

-Pico (p) is a prefix that indicates a factor of 10^{-12}

-Mili (m) is a prefix that indicates a factor of 10^{-3}

-Micro (\mu) is a prefix that indicates a factor of 10^{-6}

-Tera (T) is a prefix that indicates a factor of 10^{12}

-Kilo (K) is a prefix that indicates a factor of 10^{3}

Knowing this, let's express these quantities in terms of the SI base units:

0.13 g=0.00013 kg=1.3 (10)^{-4}kg

232 Gg=232(10)^{6}g=232 (10)^{3}kg

5.23 pm=5.23(10)^{-12}m

86.3 mg=86.3(10)^{-3}g=8.63 (10)^{-5}kg

37.6 cm=0.376 m=3.76 (10)^{-1}m

54 \mu m=54 (10)^{-6}m

1 Ts=1 (10)^{-12}s

27 ps=27 (10)^{-12}s

0.15 mK=0.15(10)^{-3}K

8 0
4 years ago
An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?
seraphim [82]
Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

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3 years ago
Explain how to measure volume using a graduated cylinder
Reil [10]
Put object into cylinder and notice the volume increased. Difference of volume is the object's volume
6 0
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Alfred Wegener believed that all of the continents were originally:
asambeis [7]
Alfred Wegener believed that all of the continents were originally attached! By looking at the map, you can see that Africa and South America look like they were once joined. This 'super continent' was known as Pangaea. He also found fossils that proved his point.
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5 0
3 years ago
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at re
xxTIMURxx [149]

Answer:

The final angular speed is 0.223 rad/s

Explanation:

By the conservation of angular moment:

ΔL=0

L₁=L₂

L₁ is the initial angular moment

L₂ is the final angular moment

L₁ is given by:

L_1=L_{door} + L_{mud}

As the door is at rest its angular moment is zero and the angular moment of mud can be considered as a point object, then:

L_1= L_{mud}= mvr

where

r is the distance from the support point to the axis of rotation (the mud hits at the center of the door; r=0.5 m)

v is the speed

m is the mass of the mud

L₂ is given by:

L_2= (I_{door} + I_{mud}) \omega_f

ωf is the final angular speed

The moment of inertia of the door can be considered as a rectangular plate:

I_{door}=\frac{1}{3}MW^2

M is the mass of the door

W is the width of the door

The moment of inertia of the mud is:

I_{mud}=mr^2

Hence,

L_1=L_2\\mvr= (I_{door} + I_{mud}) \omega_f\\\omega_f=\frac{mvr}{I_{door} + I_{mud}} \\\omega_f=\frac{mvr}{I_{door} + I_{mud}}

\omega_f=\frac{0.5kg \times 12m/s \times 0.5m}{\frac{1}{3}40kg(1m)^2+0.5kg \times (0.5m)^2}

\omega_f=0.223 \frac{rad}{s}

6 0
4 years ago
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