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belka
14 days ago
15

A cart moves along a track at a velocity of 3.5 cm/s. when a force is applied to the cart, its velocity increases to 8.2 cm/s. i

f it takes the cart 1.5 seconds to reach 8.2 cm/s, what is the acceleration of the cart? round your answer to the nearest tenth.
Physics
2 answers:
inna [987]14 days ago
5 0
In physics, acceleration describes how an object's velocity changes over time. Whenever the speed of an object shifts, it's considered to be accelerating. To calculate acceleration here:

acceleration = (8.2 - 3.5) / 1.5 = 3.1 m/s²

I trust this answers your question.
inna [987]14 days ago
5 0

According to the given definition, we need to use the formula:

vf = a * t + vo

Where,

vf: final velocity

a: acceleration

t: time duration

vo: initial velocity

From this, we isolate the acceleration variable.

The resulting formula is:

a = \frac{vf-vo}{t}

Substituting the given numbers, we get:

a = \frac{8.2-3.5}{1.5}

When rounded to one decimal place, the value becomes:

a = 3.1 \frac{cm}{s^2}

Solution:

The cart’s acceleration is:

a = 3.1 \frac{cm}{s^2}

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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
Yuliya22 [1153]

Answer: 0.10233nm

Explanation:

The mean free path of an atom can be calculated using the following equation:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is referred to as the Universal gas constant

T=0\°C=273.115K represents the absolute standard temperature

d denotes the diameter of helium atoms

N_{A}=6.0221(10)^{23}/mol symbolizes Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} indicates absolute standard pressure

<pFrom this, we can solve for d using (1), aiming to determine the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius equals half of that diameter:

r=\frac{d}{2}  (5)

Eventually:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Ultimately:

r=0.10233nm Represents the radius of the helium atom in nanometers.

5 0
10 days ago
What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?
Keith_Richards [1034]
Based on my findings, within a period of 2 hours, there are certain atoms remaining. N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0 Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0 This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms Consequently, 2.698 * 10^9 atoms represents the value of N0.
4 0
5 days ago
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i
Sav [1105]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

a=\frac{v^2}{R}

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = 9.81\ m/s^2

v = ?

R = 13.2 m

Por lo tanto,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

v = 11.4 m/s

8 0
13 days ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
Maru [1056]

Answer:

a)n= 3.125 x 10^{19 electrones.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x 10^{-3 m

radio 'r' = d/2 => 1.025 x 10^{-3 m

número de electrones 'n'= 8.5 x 10^{28}

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) La velocidad típica 'V_{d' de un electrón se expresa como:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) De acuerdo con estas ecuaciones,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

5 0
14 days ago
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A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Ostrovityanka [942]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=\frac{30-13-23*sin(45)}{3.1}

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Vf =Vi +at

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b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

5 0
7 days ago
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