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mr Goodwill [35]
3 years ago
8

Balance the following chemical equation: H3PO4 + HCl → PC15 + H20

Physics
1 answer:
kipiarov [429]3 years ago
3 0

Answer:

H3PO4 + 5HCl → PCl5 + 4H2O

Explanation:

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Who would be most likely to analyze a tablet with cuneiform to determine what goods were traded by the merchants of Ur?
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 I think your answer would Be A. Because historians mostly study written documents, while archaeologists uncover fossils and buildings.
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A 66-kg fellow stands on a digital scale in an elevator that accelerates upwards from rest to 4.5 m/s in 2.00 s. show answer Inc
Nina [5.8K]

Answer:

Explanation:

mass of the fellow ( m ) = 66kg

acceleration of fellow a

v = u + at

4.5 = 0 + a x 2

a = 4.5 /2

= 2.25 m / s²

Net force acting on fellow in upward direction by the surface of  elevator

R - mg where R is reaction force of the surface of the elevator

Applying Newton's law of motion

R - mg = ma

R = m (g +a )

= 66 x ( 9.8 + 2 )

= 778.8 N

This will be the scale reading .

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3 years ago
As the frequency of a wave decreases the wavelength
alexandr402 [8]
As frequency decreases , the wavelength will increase !!

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3 0
3 years ago
A tiger travels 3m/s^2 for 4.1s,what was its initial speed if it's final speed was 55k/h?
zaharov [31]
Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
8 0
3 years ago
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0
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