A force of 15 N acts upon a 3 kg block. Calculate the acceleration of the object.

What is the most significant change in the comet's energy as it moves from

svetoff [14.1K]

The most significant change in the comet's energy as it moves from point D to point A KE increases, GPE decreases.Option C is correct.

<h3>What are comets?</h3>

Comets are frozen leftovers from the formation of the solar system composed of dust, rock, and ice.

They range from a few miles to tens of miles wide, but as they orbit closer to the Sun, they heat up and spew gases and dust into a glowing head that can be larger than a plane.

As the comet transitions from point D to point A, the biggest energy shift is seen. GPE declines as KE rises. The right answer is C.

Hence option C is correct.

To learn more about the comet refer;

brainly.com/question/12443607

#SPJ1

3 0

2 years ago

1. True or False: Most things in motion are accelerating, few things in real life move at a constant speed.

lord [1]

1. False
2. Positive
I’m just joking I don’t know the answersss

5 0

3 years ago

Certain gases in the atmosphere trap heat on Earth.

Mamont248 [21]

The answer is <span>A.)the greenhouse effect
</span>

7 0

3 years ago

Read 2 more answers

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

3 years ago

Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo

Naily [24]

Answer:

t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

y = y₀ + $v_{oy}$ t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

0 = y₀ – ½ g t²

t= √ (2 y₀/g)

calculemos

t= √ ( 2 12 / 9,8)

t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida

x = v₀ₓ t

x = 80 1,56

x= 124,8 m

La velocidad de impacto cuando toca el suelo

vx = v₀ₓ = 80 ms

$v_{y}$ = v_{oy} – gt

v_{y} = - 9,8 1,56

v_{y} = - 15,288 m/s

la velocidad es

v = (80 i^ - 15,288 j ) m/s

Traducttion

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( $v_{oy}$ = 0)

0 =y₀I - ½ g t²

t = √ (2y₀ / g)

let's calculate

t = √ (2 12 / 9.8)

t = 1.56 s

Projectile range is the horizontal distance traveled

x = v₀ₓ t

x = 80 1.56

x = 124.8 m

Impact speed when it hits the ground

vₓ = v₀ₓ = 80 ms

v_{y} = v_{oy} - gt

v_{y} = - 9.8 1.56

v_{y} = - 15,288 m / s

the speed is

v = (80 i ^ - 15,288 j) m / s

7 0

3 years ago