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dimulka [17.4K]
3 years ago
11

What is the correct way to balance this equation? H2 + O2→H2O

Physics
2 answers:
Vsevolod [243]3 years ago
8 0

Answer:

It looks like the second one is the correct answer.

il63 [147K]3 years ago
6 0

Answer:

2H2 + O2→2H2O

Explanation:

Took the test

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State any ten reasons why students in Uganda need to study physics.(use examples of the possible technological advancements whic
Trava [24]

Answer:

The answer is "physics ".

Explanation:

Physics is the branch of science that addresses the properties of crystalline and its interaction with the fundamental elements of the universe. It covers subjects ranging in quantum mechanics with extremely little ones with quantum mechanics to the whole cosmos. You must be constant whether you like it or not, thus everyone must learn physics, irrespective of whether they're in Uganda, and plenty of other countries should have physics to dare study.

8 0
3 years ago
To make the next delivery, Sam-I-Am goes 45 mi/h for 2 hours. How far did he travel in that time?
andrezito [222]

Hello =D

This problem is about cinematic

So

V = 45 mi/h

t = 2 h

Then

V= X/t

X = V*t

Then

X = (45)*(2)

X = 90 mi

Best regards

5 0
3 years ago
A car travels 90 meters due north in 15 seconds. Then the car
tatyana61 [14]
I got you kid It’s A- 2.5m/sb
6 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
In which of these examples is the greatest movement occurring?
Elenna [48]

Answer:

not clear pic...but it's definitely not A)

7 0
3 years ago
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