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3 years ago

_____ developed the universal law of gravitation.

Physics

2 answers:

Sati [7]3 years ago

8 0

Answer: Newton

Explanation:

Anton [14]3 years ago

4 0

Sir Issac Newton developed the Universal law of Gravitation.

Explanation :

With the interesting quest for the answer why does the apply fall from a tree on the Earth, one of the wisest scientists Sir Issac Newton started investigating the action and thus developed the Universal law of Gravitation by estimating what caused the motionless apply to accelerate towards the Earth.

According to his second law of motions, an objects which are at rest (apply attached to the twig of the tree) could get in motion (fall towards the Earth) only through an external force exerted on him by the Earth. He called this force has “Gravity” or the “Gravitational Force”.

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Explain how heat transfer is occurring through convection, radiation AND conduction in this picture.

frosja888 [35]

A cup of tea/coffee transfers heat via conduction, convection and radiation.

Conduction:

Heat transfer from a coffee to the spoon is in direct contact
Inner part of the tea cup against hot liquid will transfer heat to the outer part and saucer.

Convection:

Movement of molecules within the coffee. Hot molecules of the tea at the middle of the drink will be eventually make their way to the outer part of the drink near the tea cup, where they will loose their heat.

Radiation:

Thermal radiation is the loss of heat from a solid object in to atmosphere via the release of electromagnetic radiation. So hot outer edge of the teacup loss the energy to its surroundings via thermal radiation.

4 0

2 years ago

In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mas

Aneli [31]

Answer:

a. ΔP/Δt = 42.6 N

b. F = 42.6 N

c. P = 142042.4 Pa = 1.42 KPa

Explanation:

First, we find the change in momentum of the bullets. For one bullet:

ΔP = m(Vf - Vi)

where,

ΔP = Change in Momentum = ?

m = mass of bullet = 5 x 10⁻³ kg

Vf = Final Speed = 1110 m/s

Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)

Therefore,

ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)

ΔP = 3.33 N.s

For 151 bullets:

ΔP = (151)(3.33 N.s)

ΔP = 502.83 N.s

Now, dividing this by time interval, Δt = 11.8 s

ΔP/Δt = 502.83 N.s/ 11.8 s

ΔP/Δt = 42.6 N

According to Newton's Second Law, the force is equal to rate of change of linear momentum:

Average Force = F = ΔP/Δt

F = 42.6 N

The pressure is given by:

Average Pressure = P = Average Force/Area

P = 42.6 N/ 3 x 10⁻⁴ m²

P = 142042.4 Pa = 1.42 KPa

4 0

2 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

2 years ago

Group 18 noble gases are inert because

hammer [34]

Answer:

Explanation:

have a full 8 electrons on the outermost ring

4 0

2 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

2 years ago