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Ivan
3 years ago
10

Jenny and Betty are having a great time at Busch Gardens riding the Ubanga Banga bumper cars. Jenny, who is traveling southward

in her bumper car, aims her car toward Betty, who is traveling northward in her bumper car. The cars collide and briefly come to a stop. What can you say about the magnitude of the force that Jenny's car exerts on Betty's car versus the magnitude of the force that Betty's car exerts on Jenny's car?
Physics
1 answer:
lutik1710 [3]3 years ago
7 0
They both exert an equal amount of force onto each other

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A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a
Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

Explanation:

B = Bulk modulus

Intensity, I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }

The amplitude of oscillation of the sphere is given by:

P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\

A = \triangle R\\

Substitute v and A into Pmax

P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}

I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }

P_{total} = 4\pi R^{2} I

P_{total} =4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }

The intensity of the sound wave at a distance  is given by:

I = \frac{P_{total} }{4\pi d^{2} }

I = 4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

5 0
3 years ago
a jet plane traveling 1890 km/h pulls out of a dive by moving in an arc of radius 5.20km. what is planes acceleration in g's
Andrew [12]

Answer:

53 m/s^2

Explanation:

Unit conversions:

1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s

5.2 km = 5200 m

Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

a = \frac{v^2}{r}

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

a = \frac{525^2}{5200} = 53m/s^2

7 0
3 years ago
A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
gavmur [86]
<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

5 0
3 years ago
Read 2 more answers
An object pulled to the right by two forces has an acceleration of 2.5m/s2. The free-body diagram shows the forces acting on the
Vanyuwa [196]

Answer:

490 N is the correct answer.

Explanation:

8 0
3 years ago
A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?
Anestetic [448]

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2  

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

8 0
3 years ago
Read 2 more answers
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