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3 years ago

Identify the basic structure of a chemical equation

Physics

1 answer:

guajiro [1.7K]3 years ago

6 0

The basic structure would be:

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Scientists study how the continents move. Why might scientists use a model

tigry1 [53]

Answer:

B. It is too slow to observe directly

Explanation:

They move too slow to be able to observe how they move.

I hope it helps! Have a great day!
bren~

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2 years ago

3.A bridge usually has expansion joints. They allow the bridge to become slightly longer when it

Dominik [7]

Answer:

a) When the sides of the joint are close together, the particles have more kinetic energy than they do when sides are farther apart.

Explanation:

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3 years ago

Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.

ASHA 777 [7]

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

Tension is also a force having Newton as S.I unit.

The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using Lami's theorem.

The sum of two given angles = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

$\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}$

Where

$\alpha = \beta$ = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

$\frac{T}{sin\alpha } = \frac{W}{sinY}$

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

3 0

3 years ago

When there is a Crescent Moon in the night sky, how much of the entire Moon’s surface is actually lit by the Sun? Why?

marissa [1.9K]

1 Answer. 50% of the lunar surface is always illuminated by Sun

4 0

3 years ago

Read 2 more answers

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

jeyben [28]

Initial speed of Gazelle is along x direction and its value will be

$v_x = 5.6 m/s$

also its initial height is given as

$y = 2.5 m$

Part a)

now from kinematics along Y direction

$\Delta y = v_y t + \frac{1}{2} at^2$

as we know that

$\Delta y = 0$

$v_y = 0$

$a = 9.8 m/s^2$

$2.5 = 0 + \frac{1}{2} (9.8) t^2$

$t = 0.714 s$

Part b)

distance moved horizontally

$\Delta x = v_x t$

as we know that

$v_x = 5.6 m/s$

now we will have

$v_x = 5.6 (0.714) = 4m$

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

$v_{fy} = v_y + at$

$v_{fy} = 0 + (9.8)(0.714) = 7 m/s$

$v_x = 5.6 m/s$

so net speed will be

$v^2 = v_x^2 + v_y^2$

$v^2 = 7^2 + 5.6^2$

$v = 8.96 m/s$

7 0

3 years ago