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Alex17521 [72]
2 years ago
10

an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

of her accerleration
Physics
1 answer:
Nata [24]2 years ago
7 0

The y-component of the acceleration is 0.28 m/s^2

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

Here we have:

  • Initial velocity is u=2.25 m/s at \theta_1=50.0^{\circ}, so its y-component is u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s
  • Final velocity is v=4.65 m/s at \theta_2=120.0^{\circ}, so its y-component is v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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mamaluj [8]

Explanation:

Given parameters:

Force = 30N

Weight Susan = 45kg

Weight of Dad = 100kg

Unknown:

Acceleration of Susan = ?

Acceleration of Dad = ?

Solution:

                    Force = mass x acceleration

  Acceleration = \frac{force}{mass}

 Acceleration of Susan = \frac{30}{45} = 0.67m/s²

Acceleration of Dad =  \frac{30}{100}  = 0.3m/s²

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8 0
3 years ago
How can juggling improve your hand eye coordination?​
Ne4ueva [31]

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you can predict where the juggling ball is going to land and the move you hand to catch it

Explanation:

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3 years ago
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If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/
Goshia [24]

If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)


this question is troubling me i guessed 96 ft/s

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A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
2 years ago
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