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2 years ago

14

What is the name of this atom? I

2 answers:

Vilka [71]2 years ago

5 0

Answer:

cerro azul

Explanation:

I reverse image searched and it popped up with a name.

frutty [35]2 years ago

4 0

Answer:

Helium is the name of the atom

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At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0

3 years ago

Convert 23 miles into feet.<br> **There are 5280 feet in 1 mile** *

zhenek [66]

Answer:

121440 feet is 23 miles.

5280 x 23= 121440

To check your answer just divide 121440 by 5280, which gives you 23.

Hopefully this helps you!

8 0

3 years ago

Students in physics lab times 10 oscillations of the pendulum and get 11.50 seconds how long is a pendulum in centimeters?

allochka39001 [22]

Answer: 36.5

Explanation:

7 0

2 years ago

Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to t

svetoff [14.1K]

Answer:

Explanation:

Given

rope makes an angle of $\theta$

Mass of sled and snow is m

Normal Force $=F_N$

applied Force is F

as Force is pulling in nature therefore normal reaction is given by

$F_N=mg-F\sin \theta$

Also $F\cos \theta =f_r$

$f_r=\mu _k\cdot F_N$

$f_r=\mu _k\cdot (mg-F\sin \theta )$

$F\cos \theta =\mu _kF_N$ -------1

$F\sin \theta =mg-F_N$ ---------2

Squaring 1 & 2 and then adding

$F^2=(\mu _kF_N)^2+(mg-F_N)^2$

$F=\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}$

Substitute value of F in 1

$cos\theta =\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}}$

$\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})$

8 0

2 years ago

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

8 0

2 years ago