<u>Answer:</u>
1A, 2E, 3C, 4F, 5B, 6D
<u>Explanation:</u>
Tilt meter is a an equipment used to measure very small changes from vertical level. So it is a correct match for blank space in statement A.
Richter Scale is a scale of measurement of earthquake strength, So it is a correct match for blank space in statement E.
Mercalli Intensity Scale measures effect of an earthquake, like damage caused.So it is a correct match for blank space in statement C.
Seismograph is earthquake wave measuring instrument. So it is a correct match for blank space in statement F.
Correlation Spectrometer (C.O.S.P.E.C.) is used to measure sulfur dioxide content in smoke. So it is a correct match for blank space in statement B.
Moment Magnitude Scale is an earthquake measuring scale for great earthquakes. So it is a correct match for blank space in statement D.
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
Answer:
Explanation:
Volume of block A = 10 x 6 x 1 = 60 cm³
Mass of block A = 630 g
density of mass A = mass / density
= 630 / 60 = 10.5g / cm³
Volume of block B = 5 x 5 x 3 = 75 cm³
Mass of block A = 604 g
density of mass A = mass / density
= 604 / 75 = 8.05 g / cm³
Since density of both A and B are less than that of mercury , both will float in mercury.
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
