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2 years ago

What reaction/soccur between an acid and base.

Physics

2 answers:

Elena-2011 [213]2 years ago

6 0

Answer:

neutralization reaction

Explanation:

The reaction of an acid with a base is called a neutralization reaction.

quester [9]2 years ago

4 0

es una reacción química que ocurre entre un ácido y una base produciendo sal y agua. ... Se mezcla un ácido fuerte con una base fuerte: Cuando esto sucede, la especie que quedará en disolución será la que esté en mayor cantidad respecto de la otra.

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How do you solving kinematic equations for horizontal projectiles?

daser333 [38]

See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION

7 0

2 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

8 0

2 years ago

How do resistors in parallel affect the total resistance?

4vir4ik [10]

Answer:

They're going to increase the total resistance as $R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}$

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

$I = I_1 + I_2 + ... + I_N$

where

$I_i = \frac{V_i}{R_i}$

but

$V_i = V_j = V$ for $i,j= 1, 2,..., N$

$I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}$

where

$R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}$

4 0

2 years ago

During a thunderstorm a tornado lift a car to a height of 300 m above the ground increasing and strength the tornado flings the

sasho [114]

Answer:

Explanation:

To calculate the time it took the car to hit the ground, we use the formula

speed = distance/time

80 m/s = 300 m/time

time = 300/80

time = 3.75 secs

It must have taken the car 3.75 seconds to hit the ground

To determine the horizontal distance of the car before hitting the ground, the same formula will also be used but with the time obtained above (since that was the time it took before hitting the ground)

speed = distance/time

80 = distance/3.75

distance = 3.75 x 80

distance = 300 meters

4 0

2 years ago

. A book is moved once around the perimeter of a table of dimensions 2.00 m by 3.00 m. If the book ends up at its initial positi

Lynna [10]

Answer:

Explanation:

displacement would be 10 because knowledge

5 0

3 years ago