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2 years ago

Matt (m = 80 kg) and Trey (m = 90 kg) slide across an ice skating rink and collide with each other.

Physics

2 answers:

andrew11 [14]2 years ago

7 0

(80kg)(3m/s)+(90)(-2m/s)=60N*s

Crank2 years ago

3 0

Answer:

Total momentum of the system before the collision is 60 kg-m/s.

Explanation:

Given that,

Mass of Matt, m = 80 kg

Mass of Trey, m' = 90 kg

Initial speed of the Matt before the collision, u = 3 m/s (east)

Initial speed of the Trey before the collision, u = -2 m/s (west)

After the collision, Matt is moving 1.5 m/s to the west.

We need to find the total momentum of the system before the collision. We know that the momentum is given by the product of mass and velocity. Initial momentum is given by :

$p_i=mu+m'u'\\\\p_i=80\times 3+90\times (-2)\\\\p_i=60\ kg-m/s$

So, the total momentum of the system before the collision is 60 kg-m/s. Hence, this is the required solution.

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What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u

Aleksandr-060686 [28]

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 × $10^{14}$ Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

$E=h\nu_{o}$ ..................1

here E is the energy of electron per atom $E=\frac{x}{N}$ and h is plank constant i.e. $6.626\times10^{-34} Js$ and x is binding energy

and here N is the Avogadro constant = $6.023\times10^{23}$

so E will $\frac{x}{6.023\times10^{23}}$

E = $3.19\times10^{-19} J$

so put value in equation 1 we get

$\frac{x}{6.023\times10^{23}}$ = 2.50 × $10^{14}$ × $6.626\times10^{-34} Js$

solve it we get

x = 99770.99

so binding energy is 99771 J/mol

4 0

2 years ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

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2 years ago

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A couch is pushed with a horizontal force of 80 N and moves the couch a

Lapatulllka [165]

Answer:

400 J

Explanation:

Work = force × distance

W = (80 N) (5 m)

W = 400 J

5 0

2 years ago

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A circular coil has a 10.0 cm radius and consists of 30.0 closely wound turns of wire. an externally produced magnetic field of

IrinaVladis [17]

Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:

ΦB = BA
A = πr^2
A = π(.1 m)^2
A = π/100 m^2

ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)

The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:

L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH

6 0

2 years ago

A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i

beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

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2 years ago

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