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2 years ago

An atom bonds with another atom. What is the best classification for this reaction?

1 answer:

8 0

When an atom bonds another atom, the reaction that takes place is the combination reaction. The reaction will be A + B → AB, where AB is the combined product of the reactants A and B. An example is H2 + 1/2O2 → H2O

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Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

Dimensional Analysis

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago

Which layer lies between the stratosphere and the

Pepsi [2]

The answer is Mesosphere

8 0

2 years ago

Read 2 more answers

A pendulum

RSB [31]

Explanation:

The time period rotation of the bob is t and the tension in the thread is T. ... A simple pendulum of length. ... Balancing the forces in Horizontal and vertical direction: ... Circular motion 2.

1 answer

8 0

2 years ago

A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,

Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

$v_f^2=v_o^2+2a*d$ Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

$v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}$

$v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}$

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

$15.4^2 = 30.8^2+2(-0.5)*d$

$2(0.5)*d = 30.8^2 - 15.4^2$

$d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m$

$d = 717.6 m\frac{1km}{1000m} = 0.7176Km$

8 0

2 years ago

Suppose Earth's gravitational force were decreased by half. How would this change affect a game of basketball? Write a paragraph

timama [110]

If the gravitational force were decreased by half, there would be lack of gravity on earth. Hence, it would basically affect the velocity, speed, and the distance travelled in any direction by basketball players and the ball. The basketball would bounce higher and come down in a slower speed. Whereas for the players, they would be able to leap higher from the floor.

7 0

3 years ago