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ycow [4]
3 years ago
13

1. The number of protons in an atom, unique to each element, is known as the ___________.

Physics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

atomic number

Explanation:

the number of protons in an atom is called its "atomic number"

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What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
Bas_tet [7]

Answer:

check photo for solve

Explanation:

7 0
3 years ago
Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a
Naya [18.7K]

Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

xcm = ⅜ a

Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

6 0
3 years ago
At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

Explanation:

From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

8 0
3 years ago
A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.
jeka94
  • Mass of the elevator (m) = 570 Kg
  • Acceleration = 1.5 m/s^2
  • Distance (s) = 13 m
  • Let the force be F.
  • We know, F = ma,
  • Therefore, F = (570 × 1.5) N = 855 N
  • Angle between distance and force (θ) = 0°
  • We know, work done = F s Cos θ
  • Therefore, work done by the cable during this part
  • = (855 × 13 × Cos 0°) J
  • = (855 × 13 × 1) J
  • = 11115 J

<u>Answer</u><u>:</u>

<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.5 km/h and the μs
Yakvenalex [24]

Answer:

Radius=15.773 m

Explanation:

Given data

v=29.5 km/h=8.2 m/s

μs=0.435

To find

Radius R

Solution

The acceleration is a centripetal acceleration  which is experienced by the bicycle given by

a=v^{2}/R

This acceleration is only due to static force which given as

f=ma\\f=m(v^{2}/R )

The maximum value of the static force is given as

Fs_{max}=u_{s}F_{N}

where

FN is normal force equal to mass*gravity

Therefore when the car is on the verge of sliding

f=fs_{max}\\ m(v^{2}/R )=u_{s}mg

Therefore the minimum radius should be found by the bicycle move without  sliding

So

v^{2}/R=u_{s}g\\  R=v^{2}/u_{s}g\\R=(8.2)^{2}/(0.435*9.8)\\R=15.773m

8 0
4 years ago
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