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lora16 [44]
3 years ago
15

What causes a snowflake to melt on your tongue?​

Physics
2 answers:
Elodia [21]3 years ago
7 0

Answer:

thermal energy

Explanation:

heat transfers into it causing it to physically change

frez [133]3 years ago
4 0

Answer:

A pesar de las diferencias entre cada copo de nieve y las grandes familias de copos que existen, estos todavía tienen una base hexagonal. Esta estructura hexagonal se explica por la forma en que las moléculas de agua se unen para formar un cristal

Despite the differences between each snowflake and the large families of flakes that exist, they still have a hexagonal base. This hexagonal structure is explained by the way in which water molecules come together to form a crystal.

Explanation:

mn

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A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s
BARSIC [14]

Answer:

The axial force is  P =  15.93 k N

Explanation:

From the question we are told that

     The diameter of the shaft steel is  d =  50mm

      The length of the cylindrical bushing  L =100mm

     The outer diameter of the cylindrical bushing  is  D =  70 \ mm

       The diametral interference is \delta _d = 0.005 mm

       The coefficient of friction is  \mu = 0.2

       The Young modulus of  steel is  207 *10^{3} MPa (N/mm^2)

The diametral interference is mathematically represented as

           \delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}

Where P_B is the pressure (stress) on the two object held together  

     So making P_B the subject

            P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}

Substituting values

                P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }

                 P_B = 5.069 MPa

Now he axial force required is

             P =  \mu * P_B * A

Where A is the area which is mathematically evaluated as

               \pi d l

So   P  =  \mu P_B \pi d l

Substituting values

      P =  0.2 * 5.069 * 3.142 * 50 * 100

       P =  15.93 k N

8 0
3 years ago
A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p
Luda [366]

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

F=ma

F=\dfrac{mv}{t}

F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}

F₁ = 2900 N...........(1)

F=ma

F=\dfrac{mv}{t}

F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

F=F_1+F_2

F=(2900i+(-3493.18)\ N

F=(2900i-3493.18j)\ N

Hence, this is the required solution.

6 0
3 years ago
White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w
REY [17]

Answer:

0.7515875 eV

4\times 10^{14}\leq f

Explanation:

f = Maximum frequency = 7.9\times 10^{14}\ Hz

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

W = Work function = 2.52 eV

Converting to Joules

W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J

Maximum photon energy is given by

E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J

Maximum Kinetic energy is given by

K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J

Converting to eV

1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz

The range of frequencies for which no electrons are ejected is

4\times 10^{14}\leq f

5 0
2 years ago
Which of the following is not true about tectonic plates
AlladinOne [14]

Answer:

Scientists have identified about a dozen major and several minor tectonic plates

6 0
1 year ago
A plane coming in to land at a busy airport is asked to circle the airport until the air traffic congestion eases off. The pilot
Ber [7]

Answer:

The solution is given in the picture attached below

Explanation:

8 0
3 years ago
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