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2 years ago

If a fly gets his wings cut of, is it still a fly?

Physics

2 answers:

jenyasd209 [6]2 years ago

4 0

Answer:

Explanation:

because if the fly get its wings cut off its obviously not going to be able to fly

svetoff [14.1K]2 years ago

3 0

Answer:

Yes I think... But the joke is it is a walk then.

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If you apply a force of 100 N to the level, how much force is applied to lift the crate?

Kaylis [27]

I believe that the answer is B. 133 N

4 0

3 years ago

Share your thoughts about this statement by John Wesley<br> "Electricty is the soul of universe"

ikadub [295]

Answer:

<em>I think this statement is true at least in modern day times. The world runs on nothing but technology. People use more technology instead of old school textbooks and papers. Imagine living in a world without technology… that means no cars, no trains, no devices, no machines like your stove or printer, no lights and lots more. John Wesley is 100 percent correct with this statement. Electricity is indeed the powerhouse of the universe. </em>

4 0

2 years ago

Read 2 more answers

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2

qaws [65]

Answer:

$r = 2.84 \times 10^{-14} m$

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

$\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}$

now we know that

$m = 1.67 \times 10^{-27} kg$

$e = 1.6 \times 10^{-19} C$

z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

now we have

$r = 2.84 \times 10^{-14} m$

7 0

3 years ago

An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice

Rom4ik [11]

Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

8 0

3 years ago