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denis-greek [22]
2 years ago
8

The blue mass in the image is 10kg.

Physics
2 answers:
mel-nik [20]2 years ago
8 0
I believe that the answer for this is C
dolphi86 [110]2 years ago
4 0

Answer: C

Explanation:

I guessed lol . Am i right .

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Ellus
Lesechka [4]

Answer:

W'=125.44 N

Explanation:

The weight of a person on the surface of Earth is 784 N

Weight is given by :

W = mg

m is mass of the person and g is acceleration due to gravity on the surface of Earth (10 m/s²)

m=\dfrac{W}{g}\\\\m=\dfrac{784}{10}\\\\m=78.4\ kg

The acceleration due to gravity on the surface of Moon, g' = 1.6 m/s²

Weight of the person on the moon is :

W'=mg'

W'=78.4\ kg\times 1.6\ m/s^2\\\\W'=125.44\ N

Hence, the person would weigh 125.44 N on the Moon.

6 0
2 years ago
When you push something what is that force called
shusha [124]

An

Explanation:

If you push Force is just a fancy word for pushing or pulling.

4 0
3 years ago
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori
EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

x - x_o = u_xt

\mathtt{x = u_xt  \ \  \ since (x_o = 0)}  ---- (1)

the equation of the motion y is :

\mathtt{y - y_o =u_yt - 0.5 gt^2}

\mathtt{y = u_yt-4.9t^2     \ \ \  since (y_o =0)}

\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2        }

\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}

\mathtt{1 = 5.14t - 4.9t^2}

\mathtt{4.9t^2 - 5.14t +1 = 0}

By using the quadratic formula, we have;

\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}}     }

where;

a = 4.9,   b = -5.14     c = 1

= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8}  \  \ OR \  \  \dfrac{ 5.14- \sqrt{6.8196}}{9.8}}    }

= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8}  \  \ OR \  \  \dfrac{ 5.14- 2.6114}{9.8}}    }

= \mathtt{ \dfrac{ 7.7514}{9.8}  \  \ OR \  \  \dfrac{ 2.5286}{9.8}}    }

= \mathbf{ 0.791 \  \ OR \  \  0.258}    }

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

\mathtt{x = u_x(0.258)}

\mathtt{x = ucos 40^0 (0.258)}

\mathtt{x = 8 \ cos 40^0 (0.258)}

\mathbf{x = 1.581  \ m}

Thus, the child is 1.581 m far from the fence

6 0
3 years ago
A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true o
madam [21]

Answer:

D. The motion cannot be determined without knowing the speeds of the objects before the collision.

Explanation:

This question is tricky! We know the object moving to the left has a greater mass than the one moving to the right. We'd <em>assume</em> they would move to the left because the leftwards object has a greater mass, right?

Not. So. Fast.

We can solve for the objects' final velocity using the formula for momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v .

Now here's where the trap is sprung: <em>we don't think about the equation</em>. This shows that the final velocity of the objects and the direction depends on both the mass of the objects <em>and</em> their initial velocity.

Basically, what if the 3 kg object is moving at 1 m/s and the 4 kg object is moving at –0.5 m/s? The objects would move to the <em>right</em> after the collision!

Do we know the velocity of these objects? No, right?

That means we <em>can't</em> determine the direction of their motion <u>unless we know their initial, pre-collision velocity</u>. This question is tricky because we look at the 4 kg vs. 3 kg and automatically assume the 4 kg object would dictate the direction of motion. That's not true. It depends on velocity as well.

I hope this helps you! Have a great day!

4 0
3 years ago
Please help me whit this question
mariarad [96]

Answer:

Gypsum

Explanation:

Gypsum is a soft sulfate mineral composed of Calcium Sulfate di-hydrate. It is widely used in composition of fertilizer. It is also known as land plaster. On adding gypsum to soil, the quality of soil is improved. It conditions the soil and adds nutrients. The properties of soil is improved. Hence, the correct answer is Gypsum.

Hope this helped:)

pls mark brainlist

6 0
3 years ago
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