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Kamila [148]
3 years ago
12

5. An object has a momentum of 4,000 kg-m/s and a mass of 115 kg. It crashes into another object that has a mass of 100 kg, and

the two objects stick together. If the momentum is conserved, what is the new velocity of the combined object? Round to the nearest hundredth.
A. 34.78 m/s
B. 15.82 m/s
C. 69.56 m/s
D. 18.60 m/s​
Physics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

D. 18.60

Explanation:

By the law of conservation, the momentum is neither loss nor gained but instead transfered. When they crash into each other, and stick, they combine to create a total mass of 215 kg. Since the momentum is transfered, the two objects, combined, have a total momentum of 4000 kg-m/s. We know that momentum equals mass times velocity. You then divide 4000 by 215 and get approximately 18.6 m/s

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A sailboat took 25 hours to cover 1/4 of a journey. Then, it
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Answer:

32

Explanation:

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2 years ago
Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

5 0
3 years ago
Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b
bogdanovich [222]

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

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What is the speed of a wave that has a frequency of 200 Hz and a
Luba_88 [7]
The wave speed to this question is 400 meters
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