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3 years ago

Convert 34.2 m/s into km/min. HELP

Physics

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

Explanation:

given data= 34.2 m/s

we have to convert it into km/s

as we know that

1 km=1000 m

so 1 m= 1/1000 =0.001 km

now we know that 1 min=60 sec

so 1 sec=1/60= 0.016 min

therefore

34.2m/s = 34.2*m/sec =34.2*0.001 km /0.016 min

so 34.2 m/s =2.1 km/min

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Answer:

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Explanation:

We need to determine the direction of the magnetic force. Since the velocity of the proton is in the positive x direction, and the magnetic field is in the positive y direction, we know by the vectorial formula $F=q(v\times B)$ (or, alternatively, with the left hand rule) that the magnetic force points in the positive z-direction (also taking into account that the charge is positive), so the electric field should be in the negative z-direction to balance it.

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How many volts would it take to push 1 amp through a resistance of 1 ohm?

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An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

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Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

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4 years ago

A vector of magnitude 20 is added to a vector of magnitude 25. The magnitude of this sum. might be:. A. zero. B. 3. C. 12. D. 47

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The correct answer among the choices provided is option C. The magnitude of the sum of 20 and 25 might be 12. Inequality triangle equation is used to determine if 12 is right.
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3 years ago

Electrons of wavelength 1.5 nm are emitted from a material when light of wavelength 350nm is incident on it. What is the work fu

krok68 [10]

Answer:

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For a particle with kinetic energy E and mass m , the wavelength associated is given by the following relation,

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Putting the values we get

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