Any lump of osmium is. Iridium and gold are also very close.
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
Answer:
v = 6.45 10⁻³ m / s
Explanation:
Electric force is
F = q E
Where q is the charge and E is the electric field
Let's use Newton's second law to find acceleration
F- W = m a
a = F / m - g
a = q / m E g
Let's calculate
a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8
a = 0.228 10² -9.8
a= 13.0 m / s²
Now we can use kinematics, knowing that the resting parts electrons
v² = v₀² + 2 a y
v =√ (0 + 2 13.0 1.6 10⁻⁶)
v = 6.45 10⁻³ m / s
The heat capacity and the specific heat
when we find the distance we will add all the blocks so
distance = 6+6+4
distance = 14blocks
when we find the displacement we will add and minus too
As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = <em><u>4blocks</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>west</u></em>
