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Lyrx
1 month ago
9

The equation for the change in the position of a train (measured in units of length) is given by the following expression: x = ½

at² + bt³ where a and b are constants and t is time (measured in units of time). If (mass), (length), and (time) represent units of mass, length and time respectively then b must have units of:
(A) (length)/(time)
(B) (length)/(time³)
(C) (mass)(length)
(D) (length)/(mass)
(E) (mass2)(time2)
(F) 1/(mass)(time)
(G) 1/(length)(time)
(H) (time)(mass)
(I) (mass)(time2)
(J) 1/(time³)
(K) (time)/(mass)
(L) (length)/(time²)

(mass)(length2)
Physics
2 answers:
Keith_Richards [3.2K]1 month ago
6 0
(B) (length)/(time³) Explanation: The term x = ½ at² + bt³ should meet dimensional consistency. This means that both bt³ and ½ at² must share the same units, which are length. To find the dimension of b, we rearrange the equation: [x] = [b]*[t]³ leads to length = [b]*time³, hence [b] = length/time³.
ValentinkaMS [3.4K]1 month ago
3 0
The formula representing a train's position as a function of time is given by x(t) = ½at² + bt³ where a and b are constants. We wish to identify the dimension of b. The variable x signifies length and is measured in meters, while t represents time measured in seconds. The constant b has an unknown unit, as does the constant a. With x = ½at² + bt³, we find that for dimensional correctness, units of a and b must eliminate the time units. Therefore, a should be in m/s² units. Conversely, for b to negate seconds cubed, b must be in meters/seconds³, leading to the conclusion that the unit of b is represented as meters/seconds³, which implies length/time³. Thus, the suitable answer is B.
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
2 months ago
State two advantages of a lead-acid accumulator over a leclanche cell​
kicyunya [3294]

Respuesta:

Se pueden recargar.

Poseen una vida útil mayor.

Son reutilizables.

Explicación:

Las baterías de plomo-ácido y las pilas Leclanché son tipos de células electroquímicas.

Las pilas Leclanché son células primarias. Estas células producen reacciones químicas que generan corriente eléctrica de manera irreversible.

Por su parte, las baterías de plomo-ácido son células secundarias que permiten la reversibilidad de la corriente eléctrica generada.

Esto hace que las baterías de plomo-ácido sean reutilizables, con mayor durabilidad y un tiempo de vida más prolongado.

5 0
2 months ago
A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before
Ostrovityanka [3204]

Answer:

Explanation:

Provided:

mass of the steel ball m=0.2\ kg

initial velocity of the ball u=10\ m/s

Final velocity of the ball v=-10\ m/s (moving upwards)

The impulse given is determined by the change in the momentum of the object.

<ptherefore the="" impulse="" j="" is="" defined="" by="">

J=\Delta P

\Delta P=m(v-u)

\Delta P=0.2(-10-10)

\Delta =-4\ N-s

Thus, the magnitude of the Impulse is 4 N-s.

</ptherefore>
7 0
2 months ago
Read 2 more answers
A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t
Keith_Richards [3271]

Response:

To find power, we must first determine the work done by the force.

1) We will use the following equation to calculate work:

\int\limits {F} \, dx

The force is provided by the problem; our goal is to express 'dx' in terms of 't'

2) It's known that:

\frac{dV}{dt} = a = 2.6

Thus, we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, substituting all known values gives us:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculations, the resulting work is:

161.9638 J.

4) To find power, we will use the following equation:

P = \frac{W}{t}

Thus

P = 161.9638/4.7 = 34.46 W

8 0
1 month ago
A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Yuliya22 [3333]

Response:

Clarification:

Impulse is equal to change in momentum

mv - mu, where v and u represent the final and initial velocities during the surface impact

For the downward motion of the baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

This becomes the initial velocity upon impact.

For the upward movement

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes the final velocity post-impact

change in momentum is

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

=.7056 N.s.

Impulse exerted by the floor in the upward direction is

=.7056 N.s

6 0
29 days ago
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