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2 years ago

Skater 1 has a mass of 105.0 kg and a velocity of 2.0 m/s to the left. He

Physics

1 answer:

mart [117]2 years ago

7 0

The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

<h3>What is velocity?</h3>

Velocity can be defined as the ratio of the displacement and time of a body.

To calculate the final velocity of Skater 1 we use the formula below.

Formula:

mu+MU = mv+MV............ Equation 1

Where:

m = mass of the first skater
M = mass of the second skater
u = initial velocity of the first skater
U = initial velocity of the second skater
v = final velocity of the first skater
V = final velocity of the second skater.

make v the subject of the equation.

v = (mu+MU-MV)/m................ Equation 2

Note: Let left direction represent negative and right direction represent positive.

From the question,

Given:

m = 105 kg
u = -2 m/s
M = 71 kg
U = 5 m/s
V = -3.4 m/s.

Substitute these values into equation 2

v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
v = (-210+355+241.4)/105
v = 386.4/105
v = 3.68 m/s
v ≈ 3.7 m/s

Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

Learn more about velocity here: brainly.com/question/25749514

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What are the subzones for each main zone?

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The oceanic zone is subdivided into the epipelagic, mesopelagic, and bathypelagic zones on the basis of amount of light that reaches different depths. The mesopelagic (disphotic) zone, where only small amounts of light penetrate, lies below the Epipelagic zone.

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2 years ago

1. Calculate the momentum of each car before the collision: SHOW YOUR WORK!

a_sh-v [17]

Answer:

Momentum of red car = 5kgm/s

Momentum of blue car = 0kgm/s

Explanation:

Momentum = mass × velocity

For the red car

Mass = 1kg

Velocity = 5m/s

Momentum of the red car = 1kg × 5m/s

Momentum of the red car = 5kgm/s

For the blue car.

Mass = 1kg

Velocity = 0m/s(shows that the blue car is stationery)

Momentum = 1kg ×0m/s

Momentum of the blue car = 0kgm/s

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3 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

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3 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$

Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$ .

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3 years ago

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