1. A car is slowing down at a constant rate from 40 m/s to come to a stop. The acceleration is

Which equation represents a neutralization reaction?

elixir [45]

B acid + base → water + salt

Explanation:

The best equation that represents a neutralization reaction is:

acid + base → water + salt

In a neutralization reaction, an acid reacts with a base to produce salt and water only.

Here, hydrogen combines with hydroxide to form water.

Titration is usually used to carry out this reaction in the laboratory.

An indicator is added to the base and at end the color of the reaction changes.

At this point the acids have completely been neutralized with the base.

Learn more:

Neutralization brainly.com/question/4455839

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5 0

3 years ago

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the ve

Natalka [10]

Answer:

Explanation:

mass of string = .0125 / 9.8

= 1.275 x 10⁻³ kg

Length of string l = 1.5 m .

m = mass per unit length

= ( .1.275 / 1.5) x 10⁻³ kg/m

m = .85 x 10⁻³ kg/m

wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

compare with equation of wave

y(x,t) = Acos(K x − ω t)

ω ( angular velocity ) = 4830 rad/s

k = 172 rad/m

Velocity = ω / k

= 4830/172 m /s

= 28.08 m /s

velocity of wave = $\sqrt{\frac{W}{m } }$

28.08 = $\sqrt{\frac{W}{.85\times10^{-3} } }$

788.48 = W / .85 X 10⁻³

W = 670 x 10⁻³ N .

c ) wave length

wave length =2π / k

= 2 x 3.14 / 172

= .0365 m

no of wave lengths over whole length of string

= 1.5 / .0365

= 41

d )

equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

5 0

2 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

What is one way you can destroy a magnet’s magnetism?

Alekssandra [29.7K]

-- heat the magnet red-hot in a flame
-- drop it on the floor several times, or beat it with a hammer
-- place it in a strong, rapidly alternating external magnetic field

7 0

2 years ago

A bus driver heads south with a steady speed of

Nutka1998 [239]

ANS

{ A }

Magnitude

$5400.097 \: ✓ \: m$

Direction

$25.52 \: ✓ \: ° South \: of \: West$

{ B }

Average speed in ( m/s )

$23.43 \: ✓ \: m/s$

{ C }

Magnitude

$14.06 \: ✓ \: m/s$

Direction

$25.52 \: ✓ \: ° South \: of \: West$

5 0

2 years ago

1. A car is slowing down at a constant rate from 40 m/s to come to a stop. The acceleration is - 5 m/s2. How