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3 years ago

El trabajo hecho por la gravedad durante el descenso de un proyectil es:

Physics

1 answer:

Ket [755]3 years ago

5 0

Answer:

<h2>a) positivo.</h2>

Explanation:

The question is

The work done by gravity during a projectile decline is:

a) Positive.

b) Negative.

c) Zero.

d) Depends on y-axis direction.

e) Depends on x-axis direciton.

Remember that Work is defined as the energy needed to move an object. Mathematically, it's the product between the force exerted and the displacement. That means if we don't have any displacement, then there's no work.

In this case, we are talking about the work done by the gravity force. When a projectile goes down, it's because of the gravity force, which means it can't be zero.

Specifically, the work done by gravity is positive, because when the projectile is going down, it's getting closer to the atracttive mass, which is Earth.

Therefore, the right answer is a.

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Read through the and calculate the predicted change in kinetic energy of the oblect compared to 50 kg ball traveling at 10 m/s .

Sliva [168]

Answer:

A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.

A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.

A 50 kg person falling at 10 m/s would have the same kinetic energy.

Explanation:

hope this helps:)

5 0

3 years ago

Please help. I don’t understand this

skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

v (ave) = ∆x / ∆t

and under constant acceleration,

v (ave) = (v (final) + v (initial)) / 2

According to the plot, with ∆t = 4.0 s, we have v (initial) = 0 and v (final) = 10.0 m/s, so

∆x / (4.0 s) = (10.0 m/s) / 2

∆x = ((4.0 s) • (10.0 m/s)) / 2

∆x = 20.00 m

5 0

2 years ago

Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the p

skad [1K]

The correct answer for this question is this one: "C. Neither Natalie nor Will." Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the people in the household. Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age.

7 0

3 years ago

Read 2 more answers

Every football field has two 300kg field goal posts separated by 110m of football field (that includes the endzones). What is th

Mama L [17]

Answer:

4.96×10¯¹⁰ N

Explanation:

The following data were obtained from the question:

Mass 1 (M1) = 300 Kg

Mass 2 (M2) = 300 Kg

Separating distance (r) = 110 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Gravitational force (F) =?

The gravitational force between the two goal posts can be obtained as follow:

F = GM1M2 / r²

F = 6.67×10¯¹¹ × 300 × 300 / 110²

F = 6.003×10¯⁶ / 12100

F = 4.96×10¯¹⁰ N

Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N

3 0

2 years ago

In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s

serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is s = 0.9689 m

8 0

3 years ago