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Llana [10]
3 years ago
10

Air pressure is _____.

Physics
2 answers:
fgiga [73]3 years ago
5 0

Air pressure is the weight of air on an area.  The weight of air
is due to the gravitational forces between the Earth and the
molecules of its atmosphere.

lukranit [14]3 years ago
4 0
Gravity's pull on air molecules.
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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
3 years ago
Read 2 more answers
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a
Troyanec [42]

Answer: 6.47m/s

Explanation:

The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

V = 2π * 0.309 / 0.3

V = 1.94/0.3

V = 6.47m/s

The tangential speed of the tack is 6.47m/s

7 0
3 years ago
Read 2 more answers
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
What is evaluating?
SSSSS [86.1K]

Answer:

D. Drawing a conclusion about something

3 0
3 years ago
How does cell transport make cellular respiration and photosynthesis work?
Aloiza [94]

Answer:

okjjjjkkkkjjjjhhjjikhggbvvh

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