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uysha [10]
3 years ago
15

A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue

s on at 5.8 m/s. What is her acceleration on the rough ice?
Physics
1 answer:
enot [183]3 years ago
7 0

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

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A muon is traveling at 0.988
AlexFokin [52]

As per Einstein's relation of relativity

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

here we know that

m_0 = 207 m_e = 207 \times 9.11 \times 10^{-31} kg

m_0 = 1.88 \times 10^{-28} kg

now here we know that

v = 0.988 c

now from above equation mass of the muon is given as

m = \frac{1.88 \times 10^{-28}}{\sqrt{1 - 0.988^2}}

m = 1.22 \times 10^{-27} kg

now for the momentum of muon we can use

P = mv

P = 1.22 \times 10^{-27} \times 0.988(3 \times 10^8)

P = 3.62 \times 10^{-19} kg m/s

so above is the momentum of muon

6 0
4 years ago
The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, give
Anna007 [38]

Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s

Explanation:

Given that;

mass of Earth m =  5.972 x 10²⁴ kg

radius r = 1.496 x 10¹¹ m

period t = 3.15 x 10⁷ s

now we know that Earth rotates in a circular path so the distance travelled per rotation is;

d = 2πr we substitute

d = 2π × 1.496 x 10¹¹ m

= 9.4 × 10¹¹ m

Now formula for speed v is;

v = d/t

we substitute

v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s

v = 2.98 × 10⁴ m/s

now we determine the linear momentum p

linear momentum p = mv

we substitute

p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)

p = 1.78 × 10²⁹ kg.m/s

Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s

8 0
3 years ago
Patient is in the ed due to a football hitting his nose when playing tackle football in the park. X-ray shows a displaced nasal
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ICD-10-CM codes are -S02.2XXA, W21.01XA, Y93.61, Y92.830


S02.2 for Fracture, Traumatic/Nasal (Bone(s)), ICD-10-CM Alphabetic Index. Both the open fracture code and the dislocation code are not reported. Only the fracture code is provided if a fracture and a dislocation happen at the same place. Search for "dislocation/with fracture" in the alphabetical index to be sent to a doctor. A closed fracture is a fracture with displacement. To report the conditions leading up to the injury, external cause codes are utilized. Look for Struck (accidentally) by/ball (struck) (thrown)/football W21.01 in the ICD-10-CM External Cause of Injuries Index. Seven characters are required in the Tabular List to finish the code. For the first encounter, X is utilized as a stand-in for character number six, and character number seven is given the letter A.

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6 0
2 years ago
A box is dropped from a spacecraft moving horizontally at 27.0 m/s at a distance of 155 m above the surface of a moon. The rate
gavmur [86]

Answer:

a) 10.54 sec

b) 284.58 m

c) 29.406 m/s

d) 39.92 m/s

Explanation:

Given data:

velocity of spacecraft = 27.0 m/s

rate of free fall acceleration is 2.79 m/s^2

distance of moving aircraft from mooon surface is 155 m

a. from kinematic eqaution of motion we have

y = Vi\times t + (\frac{1}{2}) a\times t^2

where y = 155 m

           Vi = 0  as this relation  is for vertical motion, so the 27.0 m/s is not included

and a = 2.79 m/s^2.

Solving for t we get

t = 10.54 sec

b.

we know that V = \frac{d}{t}

d = v\times t

   = 27 \times 10.54 = 284.58 m

c.  from the kinematic formula

v = u + at

v = 0 + 2.79\times 10.57

v = 29.4066 m/a

d. v = \sqrt { 27^2 + 29.406^2}

     v = 39.92 m/s

4 0
3 years ago
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What is the difference between speed and velocity with regards to uniform circular motion?
Mandarinka [93]
Speed ignores the direction of motion and is a scalar quantity. Velocity requires a directional component (like the arrow in a vector). In circular motion, the speed remains the same but the velocity is constantly changing since the direction of motion is constantly changing (ie. it is accelerating).
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