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3 years ago

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A beam of hydrogen molecules (h2) is directed toward a wall, at an angle of 55 with the normal to the wall. each molecule in the

beam has a speed of 1.0 km/s and a mass of 3.3 1024 g.the beam strikes the wall over an area of 2.0 cm2, at the rate of 1023 molecules per second.what is the beam’s pressure on the wall?

1 answer:

cestrela7 [59]3 years ago

8 0

The change in momentum of the particle upon hitting the wall is expressed as:

Change in momentum = Δp = 2 m v cosθ

where m = 3.3E-24 g = 3.3E-27 kg, v = 1.0 km/s = 1000 m/s, θ = 55°

Dividing both sides by Δt:

Δp / Δt = 2 (Δm / Δt) v cosθ

By definition, the force applied to a particle is equal to the change in momentum per second of the particle (by Newton's Second Law). Therefore:

Force on wall = Δp / Δt = 2 (Δm / Δt) v cosθ

We can get or calculate the value of (Δm / Δt) from the given data. That is:

Δm / Δt = m * particles per second = (3.3E-27 kg/particle) (1023 particle/s)

Δm / Δt = 3.3759 E-24 kg/s

Therefore the force is:

Total force on wall = 2 (3.3759 E-24 kg/s) (1000 m/s) cos(55)

Total Force on wall = 1.494E-22 N

Pressure = Total Force / Area = 1.494E-22 N / 2.0E-4 m^2

Pressure = 7.47E-19 Pascals

Therefore the pressure is 7.47*10^-19 Pa.

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3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

How do you express 78,000 in scientific notation?

White raven [17]

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Scientific way of thinking

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Answer:

huh,? can you explain the question more please

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3 years ago

Read 2 more answers

Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth,

Alex17521 [72]

Answer:

A) 1.67 x 10 ⁻⁶ m/s

B)5.59 x $10^-^9$ %

Explanation:

A)

Given:

d = 5.0 km,

mₐ = 2.5 x $10^1^4$ kg

u₁ = 4.0 x 10⁴ m/s

$m_n$ = 5.98 x 10 ²⁴ kg

Solve using kinetic conserved energy

mₐ x u₁ + $m_n$ x u₂ = uₓ x (mₐ + $m_n$ )

(2.5 x $10^1^4$ ) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x $10^1^4$ + 5.98 x 10 ²⁴ )

uₓ = ( 2.5 x $10^1^4$ x 4.0 x 10⁴ ) / (2.5 x $10^1^4$ + 5.98 x 10 ²⁴ )

uₓ = 1.67 x 10 ⁻⁶ m/s

B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m

t = 365 days x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s

t = 31536000 s

D = 2 π R = 2 π( 1.5 x 10 ¹¹ )

D = 9.4247 x 10 ¹¹ m

u₂ = D / t = 9.4247 x 10 ¹¹ / 31536000

u₂ = 29885.775 m/s

% = ( 1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100

% = 5.59 x $10^-^9$ %

5 0

3 years ago