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Lynna [10]
2 years ago
12

a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the

block ?​
Physics
1 answer:
Nesterboy [21]2 years ago
6 0

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

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Pavel [41]
The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.
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3 years ago
What are some common positive and negative attitudes toward physical activities? What
ra1l [238]

the common attributes are positive and negatively charged

4 0
2 years ago
Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
nignag [31]

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

           t=0 s

           t= 2 s

The ball position

          x = 40 2

          x = 80 feet

           

          y = 39 2

          y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

             x_total = 150 feet

             y _total = 80 feet

so we can see that the ball came into the field

6 0
3 years ago
A typical 12-V car battery can deliver about 750,000 C of charge before dying. This is not a lot of charge. As a comparison, cal
dimaraw [331]

Answer:

3.982 kg

Explanation:

The latent heat of vaporization = 540 cal/g

= 5.4 ×10⁵ cal/kg

L = 5.4 ×10⁵ × 4.19

  = 2.26 × 10⁶ J/kg

Q = 12 × 750,000

= 9, 000, 000

= 9 × 10⁶ J

the maximum number of kg of water (at 100 degrees Celsius) that could be boiled into steam (at 100 degrees Celsius) is:

= \frac{ 9*10^6}{2.26*10^6 }

= 3.982 kg

6 0
3 years ago
The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G
Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

7 0
3 years ago
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