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.
the common attributes are positive and negatively charged
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 +
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
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:
= 
= 3.982 kg
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