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topjm [15]
3 years ago
6

A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85

kg pin bounces backward with a velocity of 45 m/s, what is the velocity of the bowling ball after the collision
Physics
2 answers:
artcher [175]3 years ago
8 0

Answer:

v = 10.75\,\frac{m}{s}

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )

The velocity of the bowling ball after the collision is:

v = 10.75\,\frac{m}{s}

Karolina [17]3 years ago
5 0

Answer:

v1 = 10.75m/s

Explanation:

Given m = 9.0kg

u1 = 15m/s, m2 = 0.85kg, u2 = 0m/s bowling pin initially at rest. v1 =? v2 = 45m/s

Where u represents initial velocity and v final velocity for the bodies involved in the collision.

From the principle of conservation of momentum, the sum of momentum before collision is equal the sum of momentum after collision.

m1u1 + m2u2 = m1v1 + m2v2

9×15 + 0.85×0 = 9v1 + 0.85×45

135 = 9v1 +38.25

9v1 = 135 – 38.25

9v1 = 96.75

v1 = 96.75/9 = 10.75m/s

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Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
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Explanation:

5 0
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Which happened after the electromagnetic and weak nuclear forces were separated from the unified force?
alukav5142 [94]
-release of first light
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7 0
3 years ago
A spring does 80 J of work launching a 1.85 kg rock into the air. Ignoring air resistance, how high will the rock go?
Svetlanka [38]

h=80/(1.85*9.8)=4.4 m

3 0
3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

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3 years ago
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