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3 years ago

Please help!!

Physics

1 answer:

IgorLugansk [536]3 years ago

8 0

Answer:

Si un objeto se mueve en relación a un marco de referencia (por ejemplo, si una profesora se mueve a la derecha con respecto al pizarrón, o un pasajero se mueve hacia la parte trasera de un avión), entonces la posición del objeto cambia. A este cambio en la posición se le conoce como desplazamiento. La palabra desplazamiento implica que un objeto se movió, o se desplazó.

Explanation:

El desplazamiento se define como el cambio en la posición de un objeto. Se puede definir de manera matemática con la siguiente ecuación:

\text{desplazamiento}=\Delta x=x_f-x_0desplazamiento=Δx=x

−x

start text, d, e, s, p, l, a, z, a, m, i, e, n, t, o, end text, equals, delta, x, equals, x, start subscript, f, end subscript, minus, x, start subscript, 0, end subscript

x_fx

x, start subscript, f, end subscript se refiere al valor de la posición final.

x_0x

x, start subscript, 0, end subscript se refiere al valor de la posición inicial.

\Delta xΔxdelta, x es el símbolo que se usa para representar el desplazamiento.

Debemos ser cuidados al usar la palabra distancia, ya que hay dos maneras de usar el término en física. Podemos hablar acerca de la distancia entre dos puntos, o podemos hablar de la distancia recorrida por un objeto.

La distancia se define como la magnitud o el tamaño del desplazamiento entre dos posiciones. Observa que la distancia entre dos posiciones no es la misma que la distancia recorrida entre ellas.

Es importante darse cuenta que la distancia recorrida no tiene que ser igual a la magnitud del desplazamiento (es decir, la distancia entre dos puntos). De manera específica, si un objeto cambia de dirección en su trayecto, la distancia total recorrida será mayor que la magnitud del desplazamiento entre esos dos puntos. Ve los ejemplos resueltos a continuación.

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Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m

Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, $t_t$ of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + $t_t$ + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0

3 years ago

How many seconds will it take to travel 3,600 meters if your speed is 90 meters per second?

klio [65]

40

Because if you divide 3,600 by 90 it is equivalent to 40.

Hope this helps, have a blessed day :)

4 0

3 years ago

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

sweet-ann [11.9K]

Answer:

Explanation:

Given

coefficient of kinetic friction $(\mu _k)$ =0.34

inclination $\theta =44$

weight of block=51 N

(a) When block is moving upward friction force acts downward

thus

$Fsin\theta -W-f_r=0$

as block is moving with constant velocity thus $F_{net}$ is zero

$f_r=\mu _kN=0.34\times Fcos\theta$

$F\left ( \sin \theta -\mu \cos \theta \right )=W$

$F=\frac{51}{0.45}=113.31 N$

(b)When Block slides down the wall friction changes its direction to oppose the block

$Fsin\theta -W+f_r=0$

$F\left ( \sin \theta +\mu \cos \theta \right )=W$

$F=\frac{W}{\left ( \sin \theta +\mu \cos \theta \right )}$

$F=\frac{51}{0.939}=54.299 N$

3 0

3 years ago

A human subject with mass 96 kg, body specific heat 3500 Jkg^-1K^-1 skin temperature 34.5 degrees Celsius, surface area 1.5m^2 a

AlekseyPX

rate of heat radiation by the body is given by

$\frac{dQ}{dt} = \sigma e A(T^4 - T_s^4)$

here we know that

$\sigma = 5.67 \times 10^{-8}$

e = 0.7

$A = 1.5 m^2$

$T = 34.5 + 273 = 307.5 k$

$T_s = 21 + 273 = 294 k$

now from above formula rate of heat dissipation

$\frac{dQ}{dt} = (5.67 \times 10^{-8}}(0.7)(1.5)(307.5^4 - 294^4)$

$\frac{dQ}{dt} = 87.5$

now we know that

$Q = ms \Delta T$

from above equation

$\frac{dQ}{dt} = ms\frac{dT}{dt}$

now we have

$m s \frac{dT}{dt} = 87.5$

here we have

m = 96 kg

s = 3500

$96 \times 3500 \times \frac{dT}{dt} = 87.5$

$\frac{dT}{dt} = (2.6 \times 10^{-4}) \: ^0C/s$

$\frac{dT}{dt} = 0.94 ^0 C/hour$

7 0

3 years ago

Which of these is not an official SI unit of measure?

Darina [25.2K]

I believe its b. degree Celsius <span />

5 0

3 years ago