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2 years ago

The law of combining volumes only applies under these conditions EXCEPT :_________.

Physics

1 answer:

kenny6666 [7]2 years ago

3 0

Answer:

measured at constant temperature and pressure

Hope this helps :)

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Star A and star B appear equally bright in the sky. Star A is twice as far away from Earth as star B. How do the luminosities of

DerKrebs [107]

Answer:

The answer to the question is

The luminosity of stars A is four times that of star B

Explanation:

Flux (F) produced by a source of light is directly proportional to the brightness or Luminosity (L), and varies inversely to its distance d, that is $F \alpha \frac{L}{d^2}$

Therefore if the two stars present the same flux then we have

$\frac{L_1}{d_1^2} = \frac{L_2}{(2d_1)^2}$ then crossing out like terms gives $\frac{L_1}{1} = \frac{L_2}{2^2}$ or 4·L₁ = L₂

The luminosity of star A is 4 times the that of star B

7 0

3 years ago

The centers of a 15 kg lead ball and a 90 g lead ball are separated by 11 cm. part a what gravitational force does each exert on

Black_prince [1.1K]

The gravitational force between the two objects is calculated through the equation,

F = Gm₁m₂/d²

where F is the gravitational force, G is constant (equal to 6.67 x 10^-11), m₁ and m₂ are masses in kg and d is distance is meters.

Substituting the known values,

F = (6.67 x 10^-11)(15 kg)(0.090 kg)/ (0.11 m)²

F = 7.441 x 10^-9 N

ANSWER: 7.441 x 10^-9 N

8 0

3 years ago

A cheetah can run at a maximum speed

julsineya [31]

Answer:

Approximately 17.21 seconds

Explanation:

With subtraction, we have the gazelle 18.3 km/h slower than the cheetah, which is about 5.08333 m/s. As the gazelle is 87.5 meters ahead of the cheetah, 87.5 divided by 5.083333333 is about 17.21 seconds.

5 0

3 years ago

Help please i need asap

mezya [45]

Answer:

first 2 yes 3rd one no 4th one yes last one no

Explanation:

5 0

3 years ago

Read 2 more answers

a 507-kg car is in a collision with a 948-kg car. during the collision, the car applies a maximum force of 7,849 n on the truck.

katrin [286]

In a collision of 507 kg car with 948 kg truck, the maximum force that the truck applies on the car is 4208 N

Referring to the free-body diagram of each object:

Forces applied on truck

- Force from car to truck (Fct)

- Static friction force (Ft)

The net force = 0, hence,

Fct = Ft

7849 = μ . 948 . 10

Where μ is the coefficient of static friction

Hence,

μ = 7849 / 9480 = 0.83

This coefficient of static friction is assumed to be the same for the static friction force of truck and car.

Forces applied on car

- Force from truck to car (Ftc)

- Static friction force (Fc)

The net force = 0, hence,

Ftc = Fc

Ftc = μ . 507 . 10 = 0.83 . 507 . 10 = 4208 N

Learn more about friction force here:

brainly.com/question/20383236

#SPJ4

8 0

1 year ago