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jeka94
3 years ago
5

A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf. Find (a) the angle and (b) the speed vf, assuming that friction can be ignored.
Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

v_1=2.9 m/s

m_2= 72 kg

v_2=6.2 m/s

a.We have to find the angle

\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}

\theta=71^{\circ}

b. We have to find the speed v_f

According to law of conservation of momentum

m_1v_1=(m_1+m_2)v_fcos\theta

53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f

v_f=\frac{153.7}{40.7}=3.78 m/s

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sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

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3 years ago
Sound travels slower in colder air than it does in warmer air. Make a claim about the effect of air temperature on the speed of
vladimir1956 [14]

Answer: The temperature of the air affects the speed of sound in the due to the fact that colder contains air molecules with low kinetic energy. For warm air the kinetic energy of the molecules is high causing them to move in rapid motion. Because of these vibrations and collisions sound waves moves in a faster rate.

Explanation:

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3 years ago
How can the distance between the first bright band and the central band be increased in a double-slit experiment?
Strike441 [17]

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where \lambda is the wavelength of light (or any particle), D is the distance to the screen, and d is the slit separation.

From this equation we see that, by increasing the wavelength \lambda , increasing the distance from the screen D, and decreasing the slit separation d, we increase the distance between the first bright band and the central band.

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Pls help ASAP
12345 [234]

Answer:

a. 4.733 × 10⁻¹⁹ J = 2.954 eV b i. yes ii. 0.054 eV = 8.651 × 10⁻²¹ J

Explanation:

a. Find the energy of the incident photon.

The energy of the incident photon E = hc/λ where h = Planck's constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and λ = wavelength of light = 420 nm = 420 × 10⁻⁹ m

Substituting the values of the variables into the equation, we have

E = hc/λ

= 6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 420 × 10⁻⁹ m

= 19.878 × 10⁻²⁶ Jm  ÷ 420 × 10⁻⁹ m

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4.733 × 10⁻¹⁹ J = 4.733 × 10⁻¹⁹ J × 1 eV/1.602 × 10⁻¹⁹ J = 2.954 eV

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ii. What is its  maximum energy?

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= 0.054 eV

= 0.054 × 1 eV

= 0.054 × 1.602 × 10⁻¹⁹ J

= 0.08651  × 10⁻¹⁹ J

= 8.651 × 10⁻²¹ J

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Answer:

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