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Alex Ar [27]
3 years ago
6

The maximum water vapor that air can hold at 30°C is about 30 grams/cubic meter. If the humidity of the air is currently 6 grams

/cubic meter at the same temperature, the relative humidity is %.
Physics
1 answer:
zloy xaker [14]3 years ago
6 0

Answer:

20%

Explanation:

6 / 30 × 100% = 20%

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The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
Determine the energy, in electron volts, of an emitted photon when an electron transition from n=3 to n=2 occurs
tensa zangetsu [6.8K]
Give me some answer choices and i will be happy to help
7 0
3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
How much time will elapse between seeing and hearing an event?
julia-pushkina [17]

Depends on how far away the event is and what the temperature is as this affects the speed of sound.

For example, let's say you're 600 meters away and the temperature has no affect.

The speed of sound would be roughly 340 m/s so the time it would take to hear the sound would be 600/340 = 1.76 seconds

The speed of light (c) is 3.0 X 10^8 m/s so the time it would take to see the event would be 600/3 X 10^8 = 2 X 10^-7

Subtract: 1.76 - (2 X 10^-7) = approx. 1.76

3 0
3 years ago
outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
3 years ago
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