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Nikitich [7]
3 years ago
9

A body with mass 4 kg moves with a velocity of 108km/h Calculate it kinetic energy

Physics
1 answer:
Fed [463]3 years ago
4 0

Explanation:

The answer is in the pic above

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A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.900 fg (where a femtogra
just olya [345]

Answer:

9.73 x 10⁻¹⁰ m

Explanation:

According to Heisenberg uncertainty principle

Uncertainty in position x uncertainty in momentum ≥ h / 4π

Δ X x Δp ≥ h / 4π

Δp = mΔV

ΔV = Uncertainty in velocity

= 2 x 10⁻⁶ x 3 / 100

= 6 x 10⁻⁸

mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg

m = 9 x 10⁻¹⁹

Δp = mΔV

= 9 x 10⁻¹⁹ x 6 x 10⁻⁸

= 54 x 10⁻²⁷

Δ X x Δp ≥ h / 4π

Δ X x  54 x 10⁻²⁷ ≥ h / 4π

Δ X = h / 4π x 1 /  54 x 10⁻²⁷

= \frac{6.6\times10^{-34}}{4\times3.14\times54\times10^{-27}}

= 9.73 x 10⁻¹⁰ m

7 0
2 years ago
Why do plants produce their own food
Serhud [2]

Because way back, when there were some kinds of plants that made
their own food, and other kinds of plants that depended on dinosaurs or
people to bring them food, guess which ones starved and became extinct,
and which ones lived and are still among us today.


7 0
3 years ago
Read 2 more answers
You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?
Snezhnost [94]
If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.
7 0
2 years ago
What is the mass of a 3,500-N rock?
Viktor [21]
350kg because to get Newton’s it’s mass x Gravity, earths gravity is x10 so 3500 divided by 10 is 350
6 0
3 years ago
A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At
NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

B=\frac{1}{2} \mu \epsilon_0 r

Where,

\mu =Permeability constant

\epsilon_0 =Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

B \propto r

Then the distance relationship would be given by

\frac{r}{R} = \frac{B}{B_{max}}

r =\frac{B}{B_{max}} R

r = \frac{0.5B_{max}}{B_{max}}R

r = 0.5R

On the outside, however, it is defined by

B = \frac{\mu_0 i_d}{2\pi r}

Here the magnetic field is inversely proportional to the distance, that is

B \not\propto r

Then,

\frac{r}{R} = \frac{B_{max}{B}}

r = \frac{B_{max}{B}}R

r = \frac{B_{max}{0.5B_{max}}}R

r = 2R

7 0
2 years ago
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