Do Planets with more mass have more gravity than planets with less mass

1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s ,

devlian [24]

Answer:

https://youtu.be/ymHHdoCGJOU

8 0

3 years ago

0<br> What is 33 C in absolute temperature? hs

AysviL [449]

Answer:

33 Celsius is 306.15 in absolute temperature

7 0

3 years ago

A lad, waiting for his friend walks in the sidewalk, in front of her house, from the front door, first, he moves towards the Pos

Andreas93 [3]

His total displacement from his original position is -1 m

We know that total displacement of an object from a position x to a position x', d = final position - initial position.

d = x' - x

If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

x" - x' = -6 m

x" = -6 + x'

x" = -6 + 5

x" = -1 m

Thus, his total displacement from his original position is

d = final position - initial position

d = x" - x

d = -1 m - 0 m

d = -1 m

So, his total displacement from his original position is -1 m

Learn more about displacement here:

brainly.com/question/17587058

3 0

3 years ago

A golf ball reaches a height of 150 m before it stops rising and starts to fall to the ground. What is the golf balls speed (rou

artcher [175]

Answer:

v = 54 m/s

Explanation:

Given,

The maximum height of the flight of golf ball, h = 150 m

The velocity at height h, u = 0

The velocity of the golf ball right before it hits the ground, v = ?

Using the III equations of motion

Substituting the given values in the above equation,

v² = 0 + 2 x 9.8 x 150 m

= 2940

v = 54 m/s

Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s

4 0

3 years ago

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

b)

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$