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Minchanka [31]
2 years ago
14

Skater N, 58.9 kg, is moving north at a speed of 7.8 m/s when she collides with Skater E, 72.6 kg, moving east at 3.5 m/s. The t

wo skaters are stuck together. In what direction do they move after the collision?
a. 61 degrees c. 32 degrees

b. 45 degrees d. 27 degrees
Physics
1 answer:
Alik [6]2 years ago
4 0

The direction of the skaters after the collision is 61 degrees.

<h3>Momentum of the skaters</h3>

The momentum of the skaters is calculated as follows;

P(N) = 58.9 x 7.8 = 459.42 kgm/s

P(E) = 72.6 x 3.5 = 254.1 kgm/s

<h3>Direction of the skaters after collision</h3>

The direction of the skaters after the collision is calculated as follows;

\theta = tan^{-1} (\frac{459.42}{254.1} )\\\\\theta = 61^ 0

Thus, the direction of the skaters after the collision is 61 degrees.

Learn more about direction of collision here: brainly.com/question/25891473

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In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle =  35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ   with x axis    .............1

m1v3sin35 = m2v4sinθ                   with y axis  .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ  = 3.38                                            .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ  

v4 sinθ = 1.95                                              ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0
3 years ago
A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use
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The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
                           12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
                             Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107. 
7 0
3 years ago
How can I convert this?<br>Please answer with solution. Thank you.​
fomenos

Answer:

1 hr 45 min

Explanation:

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Calculate the Force of Gravity acting on an object that has a mass of 1.3 kg. The object is on Earth so use 9.8m/s2 for the acce
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To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.

Fg = m • g
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The solution is C. 12.7 N.
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3 years ago
An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 25 m lower vert
Musya8 [376]

Answer:

x = 5.79 m

Explanation:

given,

mass of the car = 39000 Kg

spring constant = 5.7 x 10⁵ N/m

acceleration due to gravity = 9.8 m/s²

height of the track = 25 m

length of spring compressed = ?

using conservation of energy

potential energy is converted into spring energy

m g h = \dfrac{1}{2}kx^2

x =\sqrt{\dfrac{2 m g h}{k}}

x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}

x =\sqrt{33.5263}

x = 5.79 m

the spring is compressed to x = 5.79 m to stop the car.

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