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Finger [1]
3 years ago
9

Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50 μm2. Convert this

to square meters.
Physics
1 answer:
Paladinen [302]3 years ago
8 0
1 m = 1 000 000 ym

converted other way we can say that:

1 ym = 10^{-6} m

Now, since we have ym^2 which is ym*ym which means:
1 ym^2 = (10^{-6}) ^2  =  10^{-12} m

we have 1,5 ym^2 which means that answer is:
1.5* 10^{-12} m
You might be interested in
Using this information...
Pepsi [2]

19.2\:\text{m/s}

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)

\:\:\:\:\:= 19.2\:\text{m/s}

6 0
3 years ago
To understand the experiment that led to the discovery of the photoelectric effect.
andrew11 [14]

Answer:

A) Emin = eV

B) Vo = (E_light - Φ) ÷ e

Explanation:

A)

Energy of electron is the product of electron charge and the applied potential difference.

The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;

Emin = eV

B)

The maximum stopping potential energy is eVo,

The energy of the electron due to the light is E_light.

If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy

Φ = E_light - eVo

Therefore,

eVo = E_light - Φ

Vo = (E_light - Φ) ÷ e

6 0
3 years ago
Which of the following is the same in all frames of reference?
Ksju [112]
The correct option is B.
The length of an object, the mass of an object and the rate of time passage for an object can change depending on the situation which the object is subject to. For instance in space, the mass and the velocity of an object usually change. But, the value of the speed of light in the space is the same for all observers regardless of the motion of an object, that is, the speed of light is a constant.<span />
4 0
3 years ago
Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?
vagabundo [1.1K]
<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

8 0
2 years ago
Read 2 more answers
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
2 years ago
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