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cluponka [151]
4 years ago
5

There are several ways to model a compound. One type of model is shown. 4 C's are connected in a line by 3 black lines. The C on

the left has 3 H's attached to it each by a black line. The next C has 1 H attached by a black line and 1 O attached by 2 black lines. The next C has 2 H's attached to it by 1 black line each. The last C has 2 H's attached to it by 1 black line each and 1 O attached by a single black line. That O has an H attached to it by 1 black line. What is the chemical formula for the molecule represented by the model? CHO C4H9O2 C4H8O C3H8O2

Physics
2 answers:
Viefleur [7K]4 years ago
8 0

Answer:

B) C4H9O2

Explanation:

Ad libitum [116K]4 years ago
5 0

Answer : The chemical formula for the molecule is C_4H_9O_2

Explanation :

Molecular formula : It is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Structural formula : It is a formula in which the lines are used between the bonded atoms and the atoms are also shown in the structural formula.

In the structural formula, the lines are used between the atoms.

As per given information of compound we conclude that, there are 4 carbon atoms, 9 hydrogen atoms and 2 oxygen atoms.

Thus, the chemical formula for the given molecule will be, C_4H_9O_2

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One student bangs two bricks together
Romashka-Z-Leto [24]

Answer:

This could be done if a stop watch is used to calculate the time taken to hear the echo and a rule should be used to calculate the distance between the bricks and the wall. Then divide distance by time

Explanation:

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7 0
4 years ago
En un barrio un estudiante observa un roedor en medio de dos postes, los cuales están comunicados por un solo cable de fibra ópt
castortr0y [4]

Answer:

7 Newton

Explanation:

Dado

Longitud de la cuerda = 50 m

El cable se dobla en 0,058 m.

Masa de roedor = 350 gramos = 0,35 kg

T = m * a + m × v2 / r

Sustituyendo los valores dados obtenemos

T = 0,35 (10 + 10)

T = 0,35 * 20

T = 7 Newton

4 0
3 years ago
A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul
romanna [79]

Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

We have the value of the angular speed, ω, in rev/min, so it is advisable to convert it to rad/sec, as follows:

ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

Now, this force is no other than centripetal force, so we can write this equation:

Ff  = Fc ⇒ μ*m*g = m*ac

⇒ μ*g = ac

Solving for μ:

μ = ac/g = 0.53 m/s² / 9.8 m/s² = 0.054

c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed  relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

So, the magnitude of the total acceleration (at the moment of maximum acceleration) as they are perpendicular each  other) , is given by the following expression:

a = √(ac)²+(at)² = 0.67 m/s²

Now, as friction force opposes to the relative movement between surfaces (the seed and the turntable), it shall be larger than the product of the mass times the total acceleration, acting along  the same action line, so we can say:

Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

5 0
3 years ago
A paleontologist measures the ratio of carbon-14 to carbon-12 in a fossil skull found at a site. What technique is he using and
pav-90 [236]
The corect anwser would be d
8 0
3 years ago
A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
4 years ago
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