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Ahat [919]
3 years ago
15

What is the net force acting on this box?

Physics
1 answer:
inn [45]3 years ago
6 0
Positive 2n to the right of the box
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Two football players are attempting to tackle each other. If one football player has a mass of 100 kg and pushes with a force of
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is it 20kg. Two opposing forces pushing onto each other

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. Write the following in the scientific notation of 0.0000002400m ​
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Explanation:

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7 0
2 years ago
Which situation would result in interference?
Arturiano [62]

Answer:

Answer is a wave increasing in energy as it hits another wave.

Explanation:

I hope it's helpful!

5 0
2 years ago
Read 2 more answers
A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0
3 years ago
A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light
SVETLANKA909090 [29]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

                                   = (3.14)(62000)²(700)

                                   = (3.14)(62)²(7)(10)⁸

                                   = 84568×10⁸

                                   = 8.45\times 10^{12} (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ 1.078\times 10^{12}

Now volume covered by each star of the galaxy = \frac{\text{Total volume of the galaxy}}{\text{Number of stars}}

= \frac{8.45\times 10^{12} }{1.078\times 10^{12}}

= 7.839 Light Years

Now the linear dimension across the volume

= (\text{Average volume per star})^{\frac{1}{3}}

= (7.839)^{\frac{1}{3}}

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

8 0
2 years ago
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