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olya-2409 [2.1K]

2 years ago

10

Hummingbirds may seem fragile, but their wings are capable of sustaining very large forces and accelerations. (Figure 1) shows d

ata for the vertical position of the wing tip of a rufous hummingbird.

1 answer:

gulaghasi [49]2 years ago

5 0

Acceleration is mathematically defined as; ΔV/t.

<h3>What is acceleration?</h3>

Acceleration is defined as the ratio of the change in velocity to time. It is defined as the extent to which the velocity changes within a given time interval.

The question is incomplete so we can not arrive at the final answer. However, acceleration is mathematically defined as; ΔV/t.

Learn more about acceleration: brainly.com/question/2437624

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A 6kg sled sliding on an icy surface experiences a 6-N frictional force exerted by the ice and the air resistive force of 0.6 N.

White raven [17]

The net force on the sled is 6.6 N pointing backwards, opposite to the direction it's sliding. That's why it's slowing down, and will eventually stop.

5 0

3 years ago

What angle is formed by the sun, the earth, and the moon during an eclipse?.

Andrew [12]

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

4 0

2 years ago

2.- Si una cámara fotográfica emite un pulso de sonido para enfocar un objeto, determinar

uranmaximum [27]

Answer:

a. El tiempo de recorrido es $5.882\times 10^{-3}$ segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo ( $t$ ), medido en segundos, a partir de la siguiente fórmula:

$t = \frac{2\cdot x_{s}}{v_{s}}$

Donde:

$x_{s}$ - Distancia entre la cámara fotográfica y el objeto, medida en metros.

$v_{s}$ - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. ( $x_{s} = 1\,m$ , $v_{s} = 340\,\frac{m}{s}$ )

$t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }$

$t = 5.882\times 10^{-3}\,s$

El tiempo de recorrido es $5.882\times 10^{-3}$ segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. ( $x_{s} = 20\,m$ , $v_{s} = 340\,\frac{m}{s}$ )

$t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }$

$t = 0.118\,s$

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

8 0

3 years ago

A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of

grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

$KE = \frac{1}{2} mv^2$

Replacing with our values we have that

$KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2$

$KE = 1.1935*10^{-13}J$

Therefore the kinetic energy of the helium nucleus is $1.1935*10^{-13}J$

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

$KE = qV$

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

$V = \frac{KE}{q}$

$V = \frac{1.19*10^{-13}}{1.6*10^{-19}}$

$V = 743750eV$

Therefore the kinetic energy in electron vols is $743750eV$

PART C) Applying the same relationship but now using the Helium core load, we will have to

$KE = QV$

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

$V = \frac{KE}{Q}$

But we need to note that the charge is equal to the number of charge for the unit charge, then

$Q = \text{No. Charge} \times \text{Unit Charge}$

$Q = (2)(1.6*10^{-19}C)$

$Q = 3.2*10^{-19}C$

Now replacing we have that

$V= \frac{1.19*10^{-13}}{32*10^{-19}}$

$V = 371875V$

Therefore the voltage applied is 371875V

5 0

3 years ago

How can you say metre cube is a derived unit

riadik2000 [5.3K]

Answer:

The newton (lowercase n!) is a derived unit because its definition consists of multiplication of three defined base units and nothing else. Its meaning is thus derived, not independently defined.

8 0

2 years ago