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3 years ago

if you knew the number of valence electrons in a nonmetal atom, how would you determine the valence of the element?

Physics

1 answer:

Vanyuwa [196]3 years ago

8 0

It would be funny because . I will not be good

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How does the kinetic energy from the forward motion of a car traveling at 16 m/s

elena55 [62]

The kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

<h3>What is Kinetic Energy?</h3>

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

For the first case, velocity v = 16m/s
For the second case, velocity = 8m/s
Let the mass of the car be m

For the first case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (16m/s)²

K = (1/2) × m × 256m²/s²

K = mass × 128m²/s²

For the second case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (8m/s)²

K = (1/2) × m × 64m²/s²

K = mass × 32m²/s²

Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

Learn more about kinetic energy here: brainly.com/question/12669551

#SPJ1

7 0

2 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

Using Newton's third law of motion, write a scientific explanation that describes why

Aneli [31]

Answer:

when a force is applied by one object to a second object, an equal and opposite force is applied back on the first object

Explanation:

4 0

2 years ago

9. When making a circuit, why is gold wiring not used?

daser333 [38]

Answer:

Gold is not used for making electric wires because it is too rare, and too costlier than Copper. Silver and copper are used for making wires. Copper is almost used in electric lines

4 0

3 years ago

to move a resting box of 100 Newton on the ground with kinetic friction coeficient of 0,250 is applied a force of 60 N horizonta

krok68 [10]

Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.

6 0

4 years ago