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Kisachek [45]
3 years ago
14

Question 4

Physics
1 answer:
Minchanka [31]3 years ago
3 0

Answer:

C

Explanation:

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momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s po
arlik [135]

Answer:

\ m/s

Explanation:

F = Force =

m = Mass of proton = 1.7\times 10^{-27\ kg

t = Time taken = 2\times 10^{-14}\ s

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2

v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s

The velocity of the proton is \ m/s

6 0
3 years ago
What happens with the plates at this Juan de fuca
djverab [1.8K]

Answer:

Subduction of the Juan de Fuca plate causes melting and magma generation in the mantle which rises to the surface to create the Cascade volcanoes.

Explanation:

8 0
3 years ago
There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in
qaws [65]

Answer:

Pressure increases due to enlargement

Explanation:

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

5 0
3 years ago
Read 2 more answers
The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0
3 years ago
What horizontally-applied force will accelerate a crate of mass 400 kg at 1 meter per second per second across a factory floor a
son4ous [18]

Answer:

The horizontally applied force = 2360 N

Explanation:

<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)

Fh = Fr + ma......... Equation 1

Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.

<em>Given: m = 400 kg, a = 1 m/s²</em>

Fr = 1/2 W, W = mg  ⇒W = 400×9.8 = 3920 N

∴Fr = 1/2(3920), Fr = 1960 N

Substituting these values into equation 1

Fh = 1960 + 400×1

Fh = 1960 + 400

Fh = 2360 N

Therefore the horizontally applied force = 2360 N

6 0
3 years ago
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