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1 year ago

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Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw

ims the first 100 meters at an average speed of 1.80 m/s how fast must he swim the second 100 meters in order to break the record?

1 answer:

Natasha_Volkova [10]1 year ago

3 0

Answer:

Explanation:

Given

average speed of Phelps $v_{avg}=2\ m/s$

total distance $d=200\ m$

he swims first 100 m at an average speed if $1.8 m/s$

so time taken is $t_1=\frac{100}{1.8}=55.55\ s$

suppose $t_2$ is the time taken to swim remaining half

average velocity is $v_{avg}=\frac{displacement}{total\ time}$

$v_{avg}=\frac{100+100}{55.55+t_2}$

$t_2+55.55=\frac{200}{2}=100$

$t_2=44.45\ s$

so velocity in the second half is

$v_2=\frac{100}{45.45}$

$v_2=2.19\approx 2.2\ m/s$

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A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

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2 years ago

Lauren wants to know which location in her apartment is best for growing African violets. She has three African violets. She put

Harrizon [31]

First, before determining which variable is which, we go over the definition of each.
The independent variable is the one which is intentionally changed in order to investigate its effect on the dependent variable.
The dependent variable is monitored and changes occur in it due to the changing conditions of the independent variable.

In this case, the location of the African violets is the independent variable as it is intentionally changed, while the rate of growth of the African violets is the dependent variable as it is being measured.

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1 year ago

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

$F_{drag}=12v+4v^2$

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

$F_{drag}=mg$

$12v+4v^2=80\times 9.8$

$4v^2+12v=784$

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

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1 year ago

Han and Greedo fire their blasters at each other. The blasts are loud, and the intensity of the sound spreads through the cantin

noname [10]

I will say it is B; the Inverse square law.
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1 year ago

Read 2 more answers

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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1 year ago