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3 years ago

There are several ways to model a compound. One type of model is shown.

Physics

2 answers:

beks73 [17]3 years ago

5 0

C4h9o2

Your welcome

Good luck

faltersainse [42]3 years ago

3 0

Answer:

What is the chemical formula for the molecule represented by the model?

$→CHO\\ →C_4H_9O_2✓ \\→C_4H_8O \\ →C_3H_8O_2$

<h3><em><u>2) C4H9O2</u></em> is the right answer.</h3>

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How high would a skater need to start on a previous incline to make it up and around a loop that is 6.1meters high?

OlgaM077 [116]

The skater need to start 7.525 meter high.

Let,

The skater need to start on a previous incline at height = h.

Mass of the skater = m

Given, height of the loop = 6.1 meter.

Radius of the loop; r = 6.1/2 meter = 3.05 meter.

Let to make it up and around a loop that is 6.1meters high, the man required minimum velocity v on the highest point of the loop.

So, centripetal force at highest point = weight of the man

⇒ mv²/r = mg

⇒ v = √(gr)

Then, potential energy of the skater at height h is = mgh.

And, minimum energy of the skater at the highest point of the loop is = potential energy + minimum kinetic energy.

= mg(2r) + 1/2 mv²

= 2mgr + 1/2 mgr

= 5/2 mgr

According to conservation of energy,

potential energy of the skater at height h = minimum energy of the skater at the highest point of the loop

⇒ mgh = 5/2 mgr

⇒ h = 5/2 r =( 5/2 )× 3.05 meter = 7.525 meter.

Hence, required height is 7.525 meter.

Learn more about energy here:

brainly.com/question/1932868

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6 0

1 year ago

In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5 minutes, and fi

lara31 [8.8K]

To develop this problem we will begin to determine the distances traveled in each of the segments using the linear motion kinematic equations. For this purpose, the distance traveled will be determined by the product between speed and time.

Part A is attached and indicates the graph of distance traveled vs time. And the calculations for development are found below. With the distance traveled and the total time it will be possible to find the average speed of the second part.

In 1.5 min the distance covered was,

$d_1 = (1.5 min)(\frac{60s}{1min}) (12m/s) = 1080m$

In the next 3.5min the distance covered is 0.

$d_2 = 0$

The distance covered for the next 2.5 min is

$d_3 = (2.5 min)(\frac{60s}{1min})(15m/s)=2250m$

Total distance covered is

$d_T = d_1+d_2+d_3$

$d_T = 1080+0+2250$

$d_T = 3330m$

For the average distance we need to use the total distance covered in the total time used. Then,

$v_{Avg} = \frac{d_T}{t_T}$

$v_{Avg} = \frac{3330m}{7.5min (\frac{60s}{1min})}$

$v_{Avg} = 7.4m/s$

6 0

3 years ago

Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 micrometers x 1.3 micrometers in area. A measuremen

lorasvet [3.4K]

Answer:

(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

(1) is correct option

Explanation:

Given that,

Thickness = 7.50 nm

Area $A=(1.3\times1.3\times10^{-6})^2$

Potential difference = 92.2 mV

Resistivity of the material $\rho=1.30\times10^{7}\ \ohm$

We need to calculate the resistance

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.30\times10^{7}\times7.50\times10^{-9}}{(1.3\times1.3\times10^{-6})^2}$

$R=3.413\times10^{10}\ \Omega$

(a). We need to calculate the amount of current that flows through this portion of the membrane

Using Ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{92.2\times10^{-3}}{3.413\times10^{10}}$

$I=2.70\times10^{-12}\ A$

The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). If the side dimensions of the membrane portion is halved

We need to calculate the new resistance

Using formula of resistivity

$R'=\dfrac{\rho \dfrac{l}{2}}{A}$

Put the value into the formula

$R'=\dfrac{1.30\times10^{7}\times\dfrac{7.50\times10^{-9}}{2}}{(1.3\times1.3\times10^{-6})^2}$

$R'=1.706\times10^{10}\ \Omega$

We need to calculate the new current

Using Ohm's law

$V=I'R'$

$I'=\dfrac{V}{R'}$

Put the value into the formula

$I'=\dfrac{92.2\times10^{-3}}{1.706\times10^{10}}$

$I'=5.404\times10^{-12}\ A$

We need to calculate the factor

$\dfrac{I'}{I}=\dfrac{5.404\times10^{-12}}{2.70\times10^{-12}}$

$I'=2I$

The factor of the current is increase by factor of 2.

Hence,(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

6 0

3 years ago

You are in a boat that is 1400 kg. The current is 1500N and is pushing you back. If you have an acceleration of 3 m/s^2 what is

aleksandrvk [35]

Answer: The force of the engines is 5700N

Explanation:

By the second Newton's law, we have that:

F = m*a

Force equals mass times acceleration.

In this case, we have:

m = 1400kg

a = 3 m/s^2

And the force will be equal to the force of the engines, f, plus the force of the current, -1500 N (because this is pushing you back, so it is in the opposite direction than f), then we have:

F = f - 1500N

Then we have the equation:

f - 1500N = 1400kg*3m/s^2 = 4200N

f - 1500N = 4200N

f = 4200N + 1500N = 5700N

The force of the engines is 5700N

5 0

3 years ago

Which terms describe the purpose of antennas on devices that use radio waves

FrozenT [24]

Answer:

The terms are

1. Transmit

2. Receive

Explanation:

What is an antenna

According to NASA

An antenna is a metallic structure that captures and/or transmits radio electromagnetic waves. Antennas come in all shapes and sizes from little ones that can be found on the roof to watch TV to really big ones that capture signals from satellites millions of miles away.

How does an antenna work?

Antennas are much more than simple devices connected to every radio. They're the transducers that convert the voltage from a transmitter into a radio signal. And they pick radio signals out of the air and convert them into a voltage for recovery in a receiver

8 0

4 years ago