All categories

4 years ago

Whose ideas did Charles Coulomb compare electric forces to?

1 answer:

3 0

He compared is to B) Isaac Newton

Isaac Newton has earlier on came up with his formula on gravitation as;

F = GMm/r²

Later on Charles Coulomb came up with his formula on electrostatic force as:

F = kQq/r²

You can see his formula is similar to that earlier formulated by Isaac Newton on gravitation.

You might be interested in

Distance Using Hubble's Law II.Find the percent difference (% = |A-O|/A × 100) between the actual (A) distance values and calcul

wlad13 [49]

Answer:

Explanation:

Let us first calculate for Virgo

$A= d_{virgo} = 17;M_{pc}, V_{virgo} = 1200 km/s,
$

Using Hubble's law

$v = H_{0}D$ For Virgo

$V_{virgo} = H_{0}D_{virgo}$

$O = D_{virgo} = \frac{v_{virgo}}{H_{0}} = \frac{1200 km/s}{70 km/s/Mpc}= 17.143 Mpc
$

Percentage difference for the Virgo

$\% = \frac{|A-O|}{A}\times 100 = \frac{|17 Mpc-17.143 Mpc|}{17 Mpc}\times 100 = 0.84 \%
$

Now for calculate for Corona Borealis

$A= d_{Corona Borealis} = 310 Mpc, v_{Corona Borealis} = 22000 km/s,
$

Using Hubble's law

$v = H_{0}D$ For Corona Borealis

$v_{Corona Borealis } = H_{0}D_{Corona Borealis }
\\O = D_{Corona Borealis } = \frac{v_{Corona Borealis }}{H_{0}} = \frac{22000 km/s}{70 km/s/Mpc}= 314.286 Mpc
$

Percentage difference for the Virgo

$\% = \frac{|A-O|}{A}\times 100 = \frac{|310 Mpc-314.286 Mpc|}{310 Mpc}\times 100 = 1.3825 \%
$

So clearly Hubble's law is more accurate for the closer objects

3 0

4 years ago

A Viking ship roller coaster at the fair has a mass of 36,000 kg. If at its peak it reaches a height of 20 m off the ground, how

adell [148]

Answer:

7056 kJ

Explanation:

Given that,

Mass of a ship roller coaster is 36,000 kg.

It reaches a height of 20 m off the ground

We need to find the gravitational potential energy does it have. The formula for the gravitational potential energy ios given by :

E = mgh

g is acceleration due to gravity

E = 36,000 kg × 9.8 m/s² × 20 m

= 7056000 J

E = 7056 kJ

So, it will have 7056 kJ of gravitational potential energy.

8 0

3 years ago

A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W

labwork [276]

Answer:

Explanation:

Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

4 0

3 years ago

A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

rewona [7]

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

4 0

4 years ago

The geologic force applied to rocks is called

amid [387]

The geologic force applied to rocks is called compression. Compression is the stress that squeezes rocks together. As a result of the compression rocks fold or fracture depending on their compressive strength or compression strength - the capacity of a material or structure to withstand loads tending to reduce size.
When the compression is horizontal the crust will be shortened and thickened. When the compression is vertical maximum a section of rock will fail in normal faults, horizontally extending and vertically thinning a given layer of rock.

4 0

3 years ago