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Vedmedyk [2.9K]
3 years ago
13

* 1a Average speed

Physics
1 answer:
ruslelena [56]3 years ago
4 0

Explanation:

\implies   v_{av} =  \dfrac{total \: displacement}{total \: time}

\implies   v_{av} =  \dfrac{100}{10}

\implies   v_{av} =10 \:  {ms}^{ - 1}

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What are the agonist and antagonist muscle of a wrist joint???
oksian1 [2.3K]
Wrist flexors are the agonist muscles, while wrist extensors are the muscle antagonists. The specific names are the flexor digitorum and the extensor digitorum.
4 0
3 years ago
Read 2 more answers
An arrow is shot upward from the roof of a building 75 m high with an initial speed of 40 m/s. If g = 10 m/s2 , what total dista
Elenna [48]

Answer:

60m

Explanation:

According to one of the equation of motions, v² = u²+2as where;

S is the distance

u is the initial velocity

v is the final velocity

a is the acceleration

Since the arrow is shot upwards, the body will experience a negative acceleration due to gravity i.e a = -g

Therefore our equation will become;

v² = u² - 2gS

Given u = 40m/s, g = 10m/s², S = 75m

Substituting to get the final velocity of the arrow we will have;

v² = 40²-2(10)(75)

v² = 1600 - 1500

v² = 100

v = √100

v = 10m/s

Total distance traveled is speed of the object × time taken

Total distance traveled = 10 × 6

= 60m

The arrow has therefore traveled 60m after 6seconds

3 0
3 years ago
A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy
snow_lady [41]

Answer:

<em> The distance required = 16.97 cm</em>

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

<em>e = 16.97 cm</em>

<em>Thus the distance required = 16.97 cm</em>

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! A IS NOT THE CORRECT ANSWER
patriot [66]

Answer: A

<u>Explanation:</u>

NOTES:

d = 650 meters

t = 10 seconds

**********************************

v = d/t

  = 650 meters/10 seconds

  = 65 meters/second

6 0
3 years ago
Read 2 more answers
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
3 years ago
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