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3 years ago

Why is Sir Francis Bacon sometimes referred to as "The Father of Modern Science"?

2 answers:

7 0

Answer: Sir Francis Bacon sometimes referred to as "The Father of Modern Science" because he proposed the scientific methods.

Explanation :

Sir Francis Bacon is sometimes referred to as "The Father of Modern Science". He is also known as the father of empiricism. He gives the knowledge about the reasoning and the events occurring in nature.

He has been also famous as "Father of Experimental Philosophy".

So, the correct option is (A) " Bacon proposed the scientific method ".

julsineya [31]3 years ago

3 0

He proposed the scientific method. He thought it was best to base a theory on observation instead of think of a theory and then try to prove it. So the answer would be A.

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Which of the following tools is used to mark the spot where a hole is to be drilled in a piece of metal?

Oxana [17]

That's a job for a center-punch.

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3 years ago

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The diagram shows organisms a diver observed during an ocean dive.

Nina [5.8K]

The diver most likely refers to the ocean's surface when describing the squid's location. Option A is correct.

<h3>What is the height?</h3>

The vertical distance between the object's top and the bottom is defined as height. It is measured in centimeters, inches, meters, and other units.

The organism is shown as;

Seaweed = - 20 meters

Clownfish = - 23 meters

Squid = - 44 meters.

The given data is a reference from the surface of the ocean. The negative sign in the data shows that the given height is below the ocean surface.

The diver most likely uses the ocean surface as a reference point to describe the position of the squid.

Hence, option A is correct.

To learn more about the height, refer to the link;

brainly.com/question/10726356

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2 years ago

A positively-charged object with a mass of 0.129 kg oscillates at the end of a spring, generating ELF (extremely low frequency)

9966 [12]

Answer:308 N/m

Explanation:

Given

mass $\left ( m\right )=0.129 kg$

wavelength $\left ( \lambda \right )=3.86\times 10^7$

We know frequency = $\frac{c}{\lambda }=\frac{3\times 10^8}{3.86\tmes 10^7}$

f=7.772 Hz

As the frequency of radio waves is same as the frequency at which object oscillates

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

$7.772=\frac{1}{2\pi }\sqrt{\frac{k}{0.129}}$

$7.772\times 2\times \pi =\sqrt{\frac{k}{0.129}}$

$k=307.70\approx 308 N/m$

7 0

3 years ago

A computer is connected across a 110 V power supply. The computer dissipates 123 W of power in the form of electromagnetic radia

nexus9112 [7]

Answer:

81.3ohms

Explanation:

Resistance is known to provide opposition to the flow of electric current in an electric circuit.

Power dissipated by the computer is expressed as;

Power = current (I) × Voltage(V)

P = IV... (1)

Note that from ohms law, V = IR

I = V/R ... (2)

Substituting equation 2 into 1, we will have;

P = (V/R)×V

P = V²/R.. (3)

Given source voltage = 100V, Power dissipated = 123W

To get resistance R of the computer, we will substitute the given value into equation 3 to have

123 = 100²/R

R = 100²/123

R = 10,000/123

R = 81.3ohms

The resistance of the computer is 81.3ohms

3 0

3 years ago

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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

3 years ago