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Goryan [66]
3 years ago
11

How do you think a graph of deceleration would differ from a graph showing acceleration

Physics
2 answers:
zloy xaker [14]3 years ago
6 0
A graph of acceleration would have to be a graph with an increasing slope, since you're going faster and faster

a graph of deceleration would have to be a graph with a negative slope since you are slowing down over time
Levart [38]3 years ago
6 0
Simply, a deceleration starts from top and goes down, and acceleration goes from bottom to top.
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Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie
Bond [772]

Answer:

Explanation:

For parallel inductors ,

\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}

\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}

L_R=16.67 mH.

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ωL_{total} where ω is angular frequency

= 2πfL_{total}

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0
3 years ago
When copper combines with oxygen to form copper oxide, the charge of the copper ion is
Archy [21]
The copper and oxygen and oxygen atoms are not ions, and the bonds are more covalent than they are ionic.
8 0
3 years ago
Read 2 more answers
The radius of our clock face is 9.2cm. It is 8:44; we are done at 9:11. How far will the minute hand (the larger one) travel?
gogolik [260]

Answer:

9:36 and how far it will travel is 26 minutes

8 0
2 years ago
AYUDAAA PORFAVOR
Sholpan [36]

Queremos crear un diagrama general para calcular el área de un triangulo.

Este será algo como:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Como naturalmente habra algunas variaciones segun el programa que utilicemos, lo voy a escribir de forma bastante general.

Primero definamos nuestras variables:

Por ejemple, en fortran usariamos algo como:

real:: B, H, A

Donde B será la variable que usaremos para la base, H para la altura, y A para el área.

Luego tenemos que escribir en pantalla algo que le diga al usario que debe introducir la base y el area.

Luego el programa debe ser capaz de leer ese input.

con algo de la forma:

B = read*input 1

H = read*input 2

Una vez tenemos definidas las variables, simplemente calculamos el área del triangulo:

A = H*B/2

Finalmente la podemos mostrar en pantalla con algo como:

print(A).

Lo que nos mostraría el valor del área.

Concluyendo, el diagrama en general sería:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Si quieres aprender más, puedes leer:

brainly.com/question/21949109

5 0
3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
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