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What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k

Molodets [167]

Answer:

$\dfrac{I_1}{I_2}=12.25$

Explanation:

$r_1$ = 14 km

$r_2$ = 49 km

Intensity of a wave is inversely proportional to distance

$I\propto \dfrac{1}{r^2}$

So,

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25$

The ratio of the intensities is $\dfrac{I_1}{I_2}=12.25$

6 0

2 years ago

When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d

Mice21 [21]

The gravitational acceleration at any distance r is given by

$g= \frac{GM}{r^2}$

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

5 0

3 years ago

Read 2 more answers

A 1569 kg car moves with a velocity of 15 m/s. What is its kinetic energy? Joules

Crank

Kinetic energy = 1/2 * mass * velocity^2

In this case,
KE = 1/2 * 1569 kg * (15 (m/s))^2 = 176,5 kN

8 0

2 years ago

At 20 C, a steel rod of length 40.000 cm and a brass rod

balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

Step 1: identify the given parameters and standard parameters

⇒Steel rod: length of rod = 40.000 cm

Young modulus(Υ) = 20 X 10¹⁰ pa

coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

stress in the rod =?

⇒Brass rod: length of rod = 30.000 cm

Young modulus(Υ) = 9 X 10¹⁰ pa

coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

stress in the rod =?

--------------------------------------------------------------------------------------------

Step 2: calculate the stress in each rod

⇒Steel rod: stress in the rod = -Y*α*ΔT

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

= -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

= -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

Step 3: calculate the new length of each rod

⇒Steel rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

New length = ΔL + L₀ = 0.0192cm + 40cm

New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

ΔL = 2400 X 10⁻⁵ cm = 0.024cm

New length = ΔL + L₀ = 0.024cm + 30cm

New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0

2 years ago

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$