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aliina [53]
3 years ago
13

a runner covers the last straight stretch of the race in 4 s. durning that time, he speeds up from 5 m/s to 9m/s what is the run

ners acceleration in this part of the race?
Physics
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

The runner's acceleration was 1\ m/s^2

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:

\displaystyle a=\frac{9-5}{4}

\displaystyle a=\frac{4}{4}=1

The runner's acceleration was 1\ m/s^2

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Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

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\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
What are the agonist and antagonist muscle of a wrist joint???
oksian1 [2.3K]
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A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
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Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

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mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

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