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3 years ago

Illustrated here are three different of carbon. They vary by the number of

Physics

1 answer:

Basile [38]3 years ago

5 0

D) isotopes; neutrons

Explanation:

The diagram illustrates the three different isotopes of carbon. They vary by the number of neutrons.

Isotopy is the existence of two or more atoms of the same element having the same atomic number but different mass numbers due to the differences in the number of neutrons in their various nuclei.

Isotopes of elements have the same electronic configuration
They share the same chemical properties
They differ in masses due the number of neutrons they contain.

Learn more:

Isotopes brainly.com/question/1915462

#learnwithBrainly

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What is the magnitude of electric field at apoint 2 meter from a point charge q= 4nc

rjkz [21]

I do not have a clue i need to answer so i can ask questions sorry

8 0

3 years ago

A runner first runs a displacement A of 3.20 km due south, and then a second displacement B that points due east. (a) The magnit

Doss [256]

Answer: 4,438.96m

Explanation:

(kindly find attachment below)

From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.

By using phythagoras theorem

H² = O² + A²

(5.38)² = (3.20)² + B²

28.944 = 10.24 + B²

B² = 28.944 - 10.24

B² = 18.7044

B = √18.7044

B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m

3 0

3 years ago

Read 2 more answers

When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the

dlinn [17]

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

$F_1 = F \frac{{343}}{(343-V)}$

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

$F_2 = F\frac{343}{(343+V)}$

We are also so told the perceived frequency decreases by 11.9%

$F_2 = F_1 - 9.27\% \text{ of } F_1$

$F_2 = F_1-0.0927F_1$

$F_2 = 0.9073F_1$

Equating,

$F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})$

$\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}$

$0.9073(343+V) = 343-V$

$(0.9073)(343)+(0.9073)V = 343-V$

$V+0.9073V = 343-(0.9073)(343)$

Solving for V,

$V = 16.67 m/s$

5 0

3 years ago

A tension force of 175 N inclined at 20.0° above the horizontal is used to pull a 40.0 - kg packing crate a distance of 6.00 m o

Murljashka [212]

Answer:

(a) Work done by the tension force is 987 J

(b) Coefficient of kinetic friction between the crate and surface is 0.495

Explanation:

(a)

Work done by any force F in moving an object by a distance d making an angle $\Theta$ with the direction of force is given by

$W=Fd\cos (\Theta )$

$W=175 \times 6 \times \cos (20 )J=987J$

Thus work done by the tension force is 987 J

(b)

Normal force on the crate is given by

$N= mg-F\sin \Theta =(40\times 9.8-175\times \sin 20)Newton$

=>N=332 Newtons

Since crate is moving with constant speed . Therefore using Newtons second law .

$Fcos(\Theta ) -\mu_k N=0$

Where $\mu_k$ =coefficient of kinetic friction

$\therefore \mu_k=\frac{F\cos (\Theta )}{N}$

=> $\mu_k=\frac{175\times \cos 20}{332}$

=> $\mu _k= 0.495$

Thus coefficient of kinetic friction between the crate and surface is 0.495

4 0

3 years ago

At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in sim

OleMash [197]

Answer:

1.869 hours

Explanation:

$T$ = Time period = 11.1 h

Angular frequency is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{11.1}\\\Rightarrow \omega=0.566\ rad/h$

The distance moved from highest to lowest level is given by

$d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d$

At the ocean surface $x_m=0.25d$

$x=x_mcos(\omega t+\phi)$

$\phi$ = Phase constant = 0 as clock is started at $x_0=x_m$

$0.25d=0.5dcos(0.56t)\\\Rightarrow cos(0.56t)=\frac{0.25}{0.5}\\\Rightarrow 0.56t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.56}\\\Rightarrow t=1.869\ h$

The time taken for the water to fall the distance is 1.869 hours

7 0

3 years ago