How do you find the instantaneous velocity on a position-time graph?

What area of science compares and describes quantities and movements of matter and energy?

qaws [65]

Answer:

the answer is physics

4 0

3 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 charge of 0.91 C is spread uniformly throughout a 25 cm rod of radius 4 mm. What are the volume and linear charge densities

Oxana [17]

The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

<h3>What is volume?:</h3>

This is the product of the height of a solid object and its crossectional area.

The Volume of the rod is can be calculated using the formula below.

Note: A rod has the shape of a cylinder.

Formula:

V = πr²h............... Equation 1

Where:

V = Volume of the rod
r = radius of the rod
h = height of the rod.

From the question,

Given:

r = 4mm = 0.004 m
h = 25 cm = 0.25 m
π = 3.14

Substitute these values into equation 1

V = 3.14(0.004²)(0.25)
V = 1.26×10⁻⁵ m³

<h3>What is linear charge density:</h3>

This is the ratio of the charge on an object to the length of the object.

The linear charge density of the rod can be calculated using the formula below.

D = Q/h.................... Equation 2

Where:

D = Linear charge density of the rod
Q = Charge on the rod.
h = height or length of the rod

From the question

Given:

Q = 0.91 C
h = 25 cm = 0.25 m

Substitute these values into equation 2

D = 0.91/0.25
D = 3.64 C/m

Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

Learn more about charge density here: brainly.com/question/14568868

4 0

2 years ago

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its

s344n2d4d5 [400]

Answer:

Part a)

Moment of inertia of the cylinder is given as

$I = 1.21 \times 10^{-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

Explanation:

As we know that the inclined plane is of length L = 3 m

and its inclination is given as 25 degree

so we know that acceleration of center of mass of the cylinder is constant so we will have

$v_f^2 = v_i^2 + 2 a L$

so we have

$v_f^2 = 0 + 2a(3)$

now we know that

$v_{avg} = \frac{L}{t} = \frac{v_f + v_i}{2}$

$\frac{3}{1.50} = \frac{v_f + 0}{2}$

$v_f = 4 m/s$

Now we have know that final speed of the cylinder due to pure rolling is given as

$v_f = \sqrt{\frac{2gH}{1 + \frac{I}{mR^2}}}$

$4 = \sqrt{\frac{2(9.81)(3 sin25)}{1 + \frac{I}{0.215(0.0319)^2}}}$

$I = 1.21 \times 10^[-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

8 0

3 years ago

Diamond has a density of 3.26 g/cm3 . what is the mass of a diamond that has a volume of 0.247 cm3 ? answer in units of g.

serg [7]

Since the fourmula is mass/volume=density multiplying the density and volume yields the mass which is <span>0.80522 grams.</span>

3 0

3 years ago