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4 years ago

Name six processes that are often involved in scientific inquiry.

1 answer:

6 0

<span>1. </span><span>Name 6 processes that are often involved in scientific inquiry.
Scientific inquiry refers to an activity of a certain person in where he or she needs to study a certain thing to be able to get knowledge and understand the scientific ideas of it on how scientists study it.
Here are the 6 process that are usually involved:
=> asking questions
=> Doing research
=> Make a Hypothesis
=> Testing the hypothesis through experiments
=> analyzing the data and draw a conclusion
=> show your result.</span>

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An object that is farther from a converging lens than its focal point always has an image that is _____.

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Smaller in size (Pt. Sized)

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3 years ago

5. You're examining some of the tiny printing on one of the newer twenty-dollar bills. A 1.5 mm tall letter appears 3 mm tall an

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Answer:

80mm or 8cm

Explanation:

According to the lens formula,

1/f = 1/u+1/v

If the object distance u = 4cm = 40mm

Object height = 1.5mm

Image height = 3mm

First, we need to get the image distance (v) using the magnification formula Magnification = image distance/object distance = Image height/object height

v/40=3/1.5

1.5v = 120

v = 120/1.5

v = 80mm

The image distance is 80mm

To get the focal length, we will substitute the image distance and the object distance in the mirror formula to have;

1/f = 1/40+1/-80

Note that the image formed by the lens is an upright image (virtual), therefore the image distance will be negative.

Also the focal length of the converging lens is positive. Our formula will become;

1/f = 1/40-1/80

1/f = 2-1/80

1/f = 1/80

f = 80mm

The focal length of the lens 80mm or 8cm

5 0

4 years ago

Read 2 more answers

you are riding on a train traveling 80 km/h. if you walk toward the back of the train at a speed of 1.2 km/h relative to the tra

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Answer:

Velocity of the passenger = 78.8[km/h]

Explanation:

In order to solve this problem, we must observe the train from a distant point of the train. In this way we can observe that the train moves at 80[km/h] relative to the ground.

In such a way that the passenger moves in a direction contrary to the movement of the train.

$v_{pass}=80-1.2\\ v_{pass}=78.8[km/h]$

The Observer located outside the train will see how the passenger moves away at 78.8 [km / h] and not at 80[km/h] which is the speed of the train.

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3 years ago

Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhan

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Answer:

84.196%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$y_0=1\ m$

x = 10 m

t = Time taken

$v_0$ = 3.5 m/s (assumed, as it is not given)

$A_0$ = $\pi 1.5^2$

We have the equation

$y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s$

$x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s$

From continuity equation we have

$Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2$

Fraction is given by

$f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%$

The fraction is 84.196%

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3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

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Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

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4 years ago