Please select the word from the list that best fits the definition

Clara and Eli used materials they could find at home to make a simple seismograph. As Clara pulls the strip of paper through the

tatyana61 [14]

Answer:

The recording that Eli and Clara's Seismograph would show will be very close to that a real Seismograph would show. This is because the mechanism they have constructed is very similar to the real one.

Explanation:

hope this helps

6 0

2 years ago

Please i need help! ill give brainliest toooo!

Igoryamba

Answer:

b

Explanation:

7 0

3 years ago

Read 2 more answers

Which of the following correctly identifies the initial direction of the force on the moving positively charged particle (i.e.,

meriva

Answer:

Explanation:

Given

Current is flowing towards right side

Considering right side i.e. +x to be positive and direction outside the paper be +k

suppose charge is below the wire and is moving towards right

$v=v_0\hat{i}$

magnetic field below the wire is outside the Paper using right hand thumb rule

$B=B_0\hat{k}$

$F=q(\vec{v}\times \vec{B})$

$F=q(v_0\hat{i}\times B_0\hat{k})$

$F=-qv_0B_0\hat{j}$

i.e. Force acts in Downward direction

i charge is moving towards left i.e.

$v=-v_0\hat{i}$

$F=q(-v_0\hat{i}\times B_0\hat{k})$

$F=qv_0B_0\hat{j}$ towards upward direction

6 0

3 years ago

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

ivolga24 [154]

Answer: $0.10233nm$

Explanation:

The mean free path $\lambda$ of an atom is given by the following formula:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is the Universal gas constant

$T=0\°C=273.115K$ is the absolute standard temperature

$d$ is the diameter of the helium atom

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ absolute standard pressure

Knowing this, let's find $d$ from (1), in order to find the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius is half the diameter:

$r=\frac{d}{2}$ (5)

Then:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

However, we were asked to find this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Finally:

$r=0.10233nm$ This is the radius of the helium atom in nanometers.

5 0

3 years ago

Whipple is confused about the connection between the velocity and acceleration of the tennis ball. he decides to compare the vel

tamaranim1 [39]

The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.

7 0

3 years ago