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jekas [21]
3 years ago
6

A student is trying to calculate the density of a ball. She already knows the mass, but she needs to determine the volume as wel

l. Which of the following formulas can be used to calculate the volume of the ball?
Physics
1 answer:
Anton [14]3 years ago
5 0
First, the student needs to get the radius of the ball. He/she can do that by simply measuring the diameter and dividing it by two to get the radius.

The next step would be to calculate the volume of the ball using the rule of volume of a perfect sphere (since the ball is considered a perfect sphere). The rule for calculating the volume of sphere is as follows:
volume of sphere = (4/3) x pi x (radius)^3 .......> this is the correct formula

The next step is to calculate the density, since the mass of the ball is known and the volume is calculated in the previous step, therefore, the density of the ball can be calculated using the following rule:
density = mass / volume
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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
Nitella [24]

Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

7 0
2 years ago
Being able to see a square in the middle of the image despite not having lines to form a square represents the Gestalt principle
vredina [299]

Answer: The correct answer for the blank is -

C. closure.

Gestalt principle of closure describes how we perceive complete figures even when the information that form the figure is missing.

This is due the fact that our brain responds to the familiar patterns inspite of getting incomplete information.

For instance, in the given question, we are able to perceive an image of square in the center despite not having actual lines that form a square.

Thus, it represents principle of closure.

7 0
3 years ago
Read 2 more answers
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
A constant power is supplied to a rotating disc .the relationship of angular velocity of disc and number of rotations made by th
Bogdan [553]
A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion. 
6 0
3 years ago
A stone on ground is zero energy​
NNADVOKAT [17]

Answer:

A stone on the ground does not have zero energy…there is an internal potential in every object. Aldo is not in action or in any mechanical motion it is being acted upon by gravity and also molecular forces and energy.

<em>Hope</em><em> </em><em>this</em><em> helps</em><em> </em><em>!</em>

7 0
3 years ago
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