The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N

Keith_Richards [3271]

Answer:

The temperature of the cooler object was nearly at room temperature. As a result, the system underwent minimal change

Explanation:

In a closed system with two objects at varying temperatures, heat energy typically flows from the hotter object to the cooler one. This transfer is more pronounced when there is a significant temperature disparity between the objects. Conversely, if the temperature difference is minor or negligible, the resulting change will be minimal.

3 0

2 months ago

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

Ostrovityanka [3204]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time $t_{sound}$ . Thus, we can derive:

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

$t_{arrow} + t _{sound} = 1 s$ .

Using this relationship in the distance formula for sound allows us to write:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Substituting the value of d from the first equation yields:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, after some calculations, we can proceed further:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Finally, the value is inserted into the initial equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

3 months ago

Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i

inna [3103]

Answer:

The surface area of the dog changes from A to 3A

Explanation:

It is stated that the dog's surface area has increased by a factor of 3 over four years.

We need to calculate the change in the relative surface area of the dog over this timeframe.

Let’s assume the initial surface area is A.

Since the surface area has been multiplied by 3,

it follows that the surface area after four years is equal to 3×A = 3A.

Thus, the dog's surface area transitions from A to 3A.

7 0

3 months ago

a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

inna [3103]

Definamos h como la distancia que hay desde el borde del pozo hasta la superficie del agua (en metros).

Consideremos la gravedad g como 9.8 m/s² y despreciemos la resistencia del aire.

La velocidad inicial vertical del guijarro es nula.
Ya que el guijarro impacta el agua tras 1.5 segundos, entonces:
h = 0.5 * (9.8 m/s²) * (1.5 s)² = 11.025 m

Resultado: 11.025 m

7 0

4 months ago

Read 2 more answers

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

kicyunya [3294]

Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s