How much force is needed to accelerate a 100 kg mass at a rate of 2.5 m/s^2

Un adolescente que va en monopatín rueda hacia abajo sobre un plano inclinado de 18.0 m de largo. El chico parte con una rapidez

qaws [65]

Answer: 31.62°

Explanation:

Tenemos como datos:

Distancia = 18.0m

Velocidad inicial = 2.0 m/s

Tiempo total = 3.3s

Sabemos que para un plano inclinado (ignorando el rozamiento) la aceleración se escribe como:

a(t) = g*sen(θ)

donde θ es el ángulo del plano inclinado, y g = 9.8m/s^2

Sabemos que para la velocidad tenemos que integrar la aceleración sobre el tiempo, entonces:

v(t) = g*sen(θ)*t + v0

Donde v0 es la velocidad inicial: v0 = 2.0m/s

v(t) = 9.8m/s^2*sen(θ)*t + 2.0m/s

Y para la posición, podemos integrar de vuelta sobre el tiempo:

p(t) = 0.5*9.8m/s^2*sen(θ)*t^2 + 2.0m/s*t + p0

Donde p0 es la posición inicial, podemos considerar que es cero para este problema.

p(t) = 4.9m/s^2*sen(θ)*t^2 + 2.0m/s*t

Y usando los datos iniciales, sabemos que en 3.3 segundos se recorren 18 metros, entonces:

p(3.3s) = 18m = 4.9m/s^2*sen(θ)*(3.3s)^2 + 2.0m/s*3.3s

18m = 51.744m*sen(θ) + 6.6m

sen(θ) = (18m - 6.6m)/ 51.744m

θ = cosec( (18m - 6.6m)/ 51.744m ) = 31.62°

4 0

3 years ago

Select the correct scientific notation form of this numeral. 106.3

Alja [10]

Scientfic notation 1.063 x 10^2

5 0

4 years ago

How much work (in J) is required to expand the volume of a pump from 0.0 L to 2.7 L against an external pressure of 1.0 atm?

spin [16.1K]

Answer:

-2.79 x 10²J

Explanation:

using the pressure volume work formula which states that

work = -PΔV

= -(1.0 atm) (2.7-0.0)L = -2.7L . atm

Convert litre.atmosphere to Joules.

1 L . atm = 101.325 joules

-2.75 L .atm = -2.75 x 101.325 = -278.64375 =

Work = -2.79 x 10²J

5 0

4 years ago

Select the correct answer.

Nimfa-mama [501]

Answer:

This makes the products have less energy than the reactants that combined to produce them.

Explanation:

6 0

3 years ago

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,

Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

$\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m$

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

$\lambda=0.004 m$ is the wavelength

So, the frequency is

$f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz$

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

$f'=\frac{v\pm v_r}{v\pm v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

$v_s$ is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

$v_r = +v$ (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

$v_s = -v$

also

v = c = speed of sound in the blood

So the formula becomes

$f'=\frac{c + v}{v - v_s}f$

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

$R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}$

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, $\rho$ .

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

3 0

3 years ago