Which of earth sphere have the most mass

Owen and Dina are at rest in frame S' , which is moving at 0.600 c with respect to frame S . They play a game of catch while Ed

Taya2010 [7]

The speed of the ball with respect to Dina is 0.800c.

A frame of reference is a set of reference points—geometric points whose positions are known mathematically and physically—that define the origin, orientation, and scale of an abstract coordinate system.

If a body does not continuously adjust its position in relation to its environment throughout the course of time, it is said to be at rest.

In frame S', Dina and Owen are at rest.

The speed of the ball with respect to Owen, u =0.800c

The speed of the fram S' with respect to frame S, v = 0.600c

Distance between Dina and Owen, L(p) = 1.8 × 10¹² m

Speed of light, c = 3 × 10⁸ m/s

Therefore, the speed of the ball according to Dina is 0.800c. As Dina and Owen are in the same frame.

Learn more about the frame of reference here:

brainly.com/question/10962551

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6 0

1 year ago

URGENT. Please help.

exis [7]

1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a

7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd

6 0

2 years ago

In which space, outdoors or in your classroom, would it be easier to hear a musician? Explain

bixtya [17]

Answer:

It is easier to hear a musician in the classroom than outdoors

Explanation:

It is easier to hear a musician in the classroom due to the improved acoustics provided by the walls of the classroom whereby along with the direct sound of the musician, which is the lead source of the sounds, there is an increased number of indirect sound reaching the ear in the classroom than outdoors and due to precedence effect, all the sound appear to come from the musician

In music played outside, along side the direct sound from the musician, the indirect sound that reach the ear is echoed from maybe by only the ground while the majority of the sound from the music wanders away with the wind and in other directions as well as being absorbed such that speakers will be required to improve the sound of the music outdoors.

7 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

2 years ago