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Juliette [100K]

3 years ago

11

YAHOO ANSWERSA 1260 kg car moving south at 12.9 m/s collides with a 2430 kg car moving north. The cars stick together and move a

s a unit after the collision at a velocity of 4.38 m/s to the north. Find the velocity of the 2430 kg car before the collision.

1 answer:

Fofino [41]3 years ago

7 0

Explanation:

Below is an attachment containing the solution.

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A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two

stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

$T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s$

FOR THE WAVE TRAVELING THROUGH AIR:

$T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s$

The separation in time between two pulses can now be given as follows:

$\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\$

<u>ΔT = 0.02412 s</u>

3 0

3 years ago

In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m

ahrayia [7]

Answer:

The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²

Explanation:

Given;

density of the liquid, ρ = 1500 kg/m³

frequency of the wave, F = 410 Hz

wavelength of the sound, λ = 7.80 m

The speed of the wave is calculated as;

v = Fλ

v = 410 x 7.8

v = 3,198 m/s

The bulk modulus of the liquid is calculated as;

$V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2$

Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²

5 0

3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

6 0

3 years ago

Read 2 more answers

What is the accletation of a 1500kg with a net force of 7500 N

matrenka [14]

Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.

3 0

3 years ago

A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i

Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

5 0

4 years ago