Refer to your lab guide and review the data in Table F to complete each statement.

A rock is thrown straight up into the air. At the

Aleonysh [2.5K]

The answer is 3.

When a ball is thrown into the air, there is no friction (there's nothing rubbing against it), so the only force acting on the ball is gravity, which aims to pull the ball back down. Since this is the only force, the answer will equal the gravity acting on the ball. Since "weight" is defined as just that (the gravity acting on something), the answer is 3, the magnitude of the rock's weight (or the force of gravity).

6 0

4 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

KATRIN_1 [288]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

3 0

3 years ago

Which of the following is an accurate statement?

weeeeeb [17]

An object's momentum will change if there is a non-zero net external force acting on it. This assertion is true.

<h3></h3><h3>What is momentum?</h3>

The momentum is defined as the product of mass and the velocity of the body. It is denoted by the letter P. It occurs due to the applied force. Its unit is Kg m/s².

p=mΔV

If there is a change in the velocity there must be a force acting on the object.

If an object is acted on by a non-zero net external force, its momentum will not remain constant. is an accurate statement.

Hence, option B is correct.

To learn more about the momentum refer to the link;

brainly.com/question/4956182

#SPJ1

7 0

2 years ago

A car travels 50 miles in 2.5 hours. Calculate the average velocity

Dahasolnce [82]

In order to calculate velocity, we would need to know where the car was
when it started out, and where it was when it stopped. Without any of this
information, the best we can do is calculate its average speed.

Average speed = (distance covered) / (time to cover the distance)

   = (50 miles)    / (2.5 hours)

   =    (50 / 2.5)    mile/hour

   = 20 mile/hour .

5 0

4 years ago

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

3 0

3 years ago