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alex41 [277]
2 years ago
7

How far out from the sun is the rock line?

Physics
1 answer:
Vesna [10]2 years ago
8 0

Answer:

5 million miles

Explanation:

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If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p
Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0
2 years ago
Which of the following is true about producers?
Nina [5.8K]

the answer is d they are essential to all ecosystems
4 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3
Inga [223]
m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.

The momentul of the system preserves:

m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with E.

Now, we write the energy conservation law:

\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E

From the above equation, you find E,  and then conclude that the sound energy can certainly not be greater than this.
8 0
3 years ago
Exercise can help prevent?
Sedaia [141]

health conditions and diseases


8 0
3 years ago
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