What are 3 artificial and 2 natural sources of electromagnetic radiation?

Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?

ValentinkaMS [17]

Answer:

B. The object's volume

Explanation:

When an object is immersed in a fluid, it experiences an upward force which is called buoyant force. The magnitude of the buoyant force is given by:

$B=\rho_f V_{disp} g$

where

$\rho_f$ is the density of the fluid in which the object is immersed

$V_{disp}$ is the volume of the fluid displaced by the object

$g$ is the acceleration due to gravity

When the object is totally immersed in the fluid, $V_{disp}$ corresponds to the volume of the object; when the object is only partially immersed, $V_{disp}$ corresponds only to the volume of the part of the object immersed.

From the formula, we see that the greatest buoyant force is experienced by the object when it is fully immersed. Moreover, we see that the buoyant force depends only on one property of the object: its volume. Therefore, the correct choice is

B. The object's volume

6 0

3 years ago

A car races around a circular track. Friction on the tires is the what that acts toward the center of the circle and keeps the c

Ivanshal [37]

I would think force. Because friction has to have force to work ^-^

8 0

3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

A basketball player does 2.43 x 105 J of work during her time in the game, and evaporates 0,1 '10 kg of water. Assuming a latent

olchik [2.2K]

The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal

For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.

The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.

From the given information,

the work done on the basketball is dW = 2.43 × 10⁵ J

The amount of heat loss is represented by dQ.

where;

dQ = -mL

∴

Using the first law of thermodynamics:b

dU = dQ - dW

dU = -mL - dW

dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)

dU = -491.6 × 10³ J

dU = -491.6 kJ

The number of nutritional calories the player has converted to work and heat can be determined by using the relation:

$\mathbf{dU = -491.6 \ kJ \times (\dfrac{1 \ cal}{ 4.186 \ J})}$

dU = -117.44 kcal

Learn more about first law of thermodynamics here:

brainly.com/question/3808473?referrer=searchResults

5 0

2 years ago

Help it’s due tomorrow

Sati [7]

Answer:

A., the variables have a direct relationship.

Explanation:

As K rises, L rises.

It's not B. because one isn't rising as the other is lowering.

It's not C. because undefined would be a vertical line.

4 0

3 years ago