It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

$v^{2}-u^{2}= 2as$

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$v^{2}-(0)^{2}= 2(1.44)(30)$

$v= 9.3 m^{-1}$

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

$s=ut+\frac{(1)}{2}at^2$

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$30=(0)t+\frac{(1)}{2}(1.44)t^2$

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

$T=6.45+\frac{s'}{v}$

Put s= 30 m, $v= 9.3 m^{-1}$ and s'= 100 m.

$T=6.45+\frac{(100)}{9.3}$

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

6 0

3 years ago

Write the following number in scientific notation 156.60

Liula [17]

Answer:

1.566 x 10^2

Move the decimal to where the number being multiplied by 10^x is greater than 1 but less than 10. Then multiply it by 10^x

X is the number of times you moved the decimal, so in this case it would be 10^2

8 0

3 years ago

A stone is tied to a 0.85-meter cord. it is swung in a circle at a constant rate of 6.0 m/s. what is the centripetal acceleratio

Andrei [34K]

Answer:42.4m/s^2

Explanation:

Velocity(v)=6m/s

Radius(r)=0.85 meter

Centripetal acceleration=(v x v) ➗ r

Centripetal acceleration=(6 x 6) ➗ 0.85

Centripetal acceleration=36 ➗ 0.85

Centripetal acceleration=42.4

5 0

3 years ago

Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the

grandymaker [24]

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

3 0

3 years ago

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