All categories

3 years ago

The scientific method is

2 answers:

8 0

The scientific method is a body of techniques for investigating phenomena, acquiring new knowledge, or correcting and integrating previous knowledge. To be termed scientific, a method of inquiry is commonly based on empirical or measurable evidence subject to specific principles of reasoning.

lakkis [162]3 years ago

7 0

A method of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.

You might be interested in

Which lists the amplitudes of sound waves from these sources in order from greatest to least?

nignag [31]

The correct option is:

chainsaw, diesel truck, rustling leaves

In fact, this option lists the sounds from greatest amplitude (which means, greatest loudness) to smallest amplitude. For comparison, this is the intensity of the three different sounds:
- chain saw: approximately 120 dB
- diesel truck: approximately 80 dB
- rustling leaves: approximately 20 dB

(values taken from this table: https://www.chem.purdue.edu/chemsafety/Training/PPETrain/dblevels.htm )

5 0

3 years ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Reil [10]

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

6 0

3 years ago

Read 2 more answers

PWEESE HELP ME WIT MY QUIZ

kaheart [24]

I believe it is acceleration

7 0

2 years ago

Read 2 more answers

Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

3 years ago