In which position are the earth, moon, and sun during a full moon?

2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is

larisa86 [58]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

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HELP PLS DUE SOON DUE SOON NEED HELP PLS PLS I BEG U

Bumek [7]

Answer:

The thirds option

Explanation:

3 0

2 years ago

If a tank contains water 4 m deep, what is the pressure at the bottom of the tank? (Neglect the atmospheric pressure.)

makkiz [27]

Given:

Water 4 m deep

Required:

Pressure at the bottom of the tank

Solution:

p2 – p1 = gh

p2 – p1 = p = gh

p = gh = 1000kg/m3 (9.8m/s2)(4m)

6 0

3 years ago

A)A certain medical machine emits x-rays with a minimum wavelength of 0.024 nm. One day, the machine has an electrical problem a

oee [108]

Answer:

(a) 0.032 nm

(b) 39,235 eV

(c) 70,267.8 eV

Explanation:

(a) The energy of a photon can be calculated using:

E = hc/λ equation (1)

where:

h = 4.13*10^-15 eV.s

c = 3*10^8 m/s

λ = 0.024*10^-9 m

Thus:

E = (4.13*10^-15)*(3*10^8)/0.024*10^-9 = 51,625 eV

Then we calculate 76% of this estimated energy and determine the new wavelength:

$E_{new} = 0.76*51625 = 39,235 eV$

Using equation (1) to determine the new wavelength:

λ $_{new} = \frac{h*c}{E_{new} }$

λ $_{new}$ = (4.13*10^-15)*(3*10^8)/39235 = 3.15*10^-11 m = 0.032 nm

(b) As calculated in part (a), the maximum x-ray energy this machine can produce is $E_{new} = 0.76*51625 = 39,235 eV$

(c) The energy of a Ka x-ray photon can be estimated using:

$E_{ka} = (10.2 eV)*(Z-1)^{2}$

where Z is the atomic number = 84.

$E_{ka} = (10.2 eV)*(84-1)^{2}$ = 70,267.8 eV

4 0

3 years ago

A proton moves with a speed of 4.00 106 m/s horizontally, at a right angle to a magnetic field. What magnetic field strength is

Free_Kalibri [48]

Answer:

Magnetic field, $B=2.55\times 10^{-14}\ T$

Explanation:

It is given that,

Speed of proton, $v=4\times 10^6\ m/s$

Mass of the proton, $m=1.67\times 10^{-27}\ kg$

Charge on proton, $q=1.6\times 10^{-19}\ C$

We need to find the magnetic field strength required to just balance the weight of the proton and keep it moving horizontally.

The Lorentz force is given by :

$F=q(v\times B)=qvB\ sin90$ .............(1)

The weight of proton,

$W=mg$ ..............(2)

From equation (1) and (2), we get :

$mg=qvB$

$B=\dfrac{mg}{qv}$

$B=\dfrac{1.67\times 10^{-27}\ kg\times 9.8\ m/s^2}{1.6\times 10^{-19}\ C\times 4\times 10^6\ m/s}$

$B=2.55\times 10^{-14}\ T$

Hence, this is the required solution.

3 0

2 years ago