A projectile is launched at an angel into the air its verticle acceleration is g?

Kinetic energy is the energy of motion and increases with the speed of the wave.

hammer [34]

Kinetic energy is movement so A
Potential is still so false

5 0

2 years ago

Se lanza una piedra de 3.00 N verticalmente hacia arriba desde el suelo. Se observa que, cuando está 15.0 m sobre el suelo, viaj

BARSIC [14]

Answer:

(a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

Explanation:

Given that,

Weight of stone = 3.00 N

Height = 15 m

Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

$W=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

Put the value into the formula

$-9.8\times15=\dfrac{1}{2}\times(v_{2}^2-v_{1}^2)$

$-2\times9.8\times15=25^2-v_{1}^2$

$-v_{1}^2=-300-25^2$

$v_{1}=\sqrt{925}$

$v_{1}=30.41\ m/s$

(b). We need to calculate the maximum height

Using work energy theorem

$[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

$mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

Here, $\dfrac{1}{2}mv_{2}^2$ =0

$-(mg)\times d=\dfrac{1}{2}mv_{1}^2$

$d=\dfrac{v_{1}^2}{2g}$

Put the value into the formula

$d=\dfrac{30.41^2}{2\times9.8}$

$d=47.18\ m$

Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

6 0

2 years ago

Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 2.0 104 kg (about 20 tons). The ro

elena-14-01-66 [18.8K]

Answer:

r = <2.640 10⁶, 1.01 10⁸, 0> m

Explanation:

For this exercise we are going to solve it for each direction separately,

we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is

r₀ = <4, 4, 0> 10³ m

X axis

the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off

They ask where it is after one hour t = 1 h = 3600 s

Let's apply Newton's second law

Fₓ = m aₓ

aₓ = Fₓ / m

aₓ = 7.0 10⁵/2 10⁴

aₓ = 3.5 10¹ m / s

Let's use kinematics to find the distance

for the first t₁ = 21 s the movement is accelerated

x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²

x₁ = x₀ + ½ aₓ t₁²

x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5

x₁ = 11717.5 m

this instant has a speed of

vₓ = v₀ₓ + aₓ t

vₓ = aₓ t ₁

vₓ = 35 21

vₓ = 735 m / s

the rest of the time there is no acceleration so it is uniform motion at this speed

t₂ = 3600 - 21

t₂ = 3576 s

vₓ = x₂ / t₂

x₂ = vₓ t₂

x₂ = 735 3576

x₂ = 2,628 10⁶ m

the total distance traveled in this direction is

x_total = x₁ + x₂

x_total = 11717.5 + 2.628 10⁶

x_total = 2,640 10⁶ m

Y axis

on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of $v_{oy}$ = 28 10³ m / s and there is no force on this axis F_{y} = 0

The movement in this axis is uniform,

v_{y} = y / t

y = v_{y} t

y = 28 10³/3600

y = 1.01 10⁸ m

the total distance is

y_total = y₀ + y

y_total = 4000 + 1.01 10⁸

y_total = 1.01 10⁸ m

Z axis

the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0

the movement is uniform

z = 0

the final position of the rocket after 1 h is

r = <2.640 10⁶, 1.01 10⁸, 0> m

7 0

3 years ago

Pls help me fast and correct

bekas [8.4K]

Answer:

A

Explanation:

due to position at height is the answer

4 0

1 year ago

Mrs. Miller is acting a little strange and is pushing as hard as she can against a wall. Using Newton's Laws, what could be say

deff fn [24]

Newton’s 3rd Law of Motion

8 0

2 years ago

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