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3 years ago

5

In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times

in 60 seconds. The magnitude of the player's average velocity during the drill is...
1. 0 m/s
2. 0.5 m/s
3. 1 m/s
4. 1.5 m/s
5. 3 m/s
6. 30 m/s

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

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Name the stages of a volcano and name the three different types of volcano mountains caused by the lava buildup.

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3 years ago

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Which of the statements best describes density?

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Answer:

The answer to your question is: letter D.

Explanation:

a.The mass that a mole of substance has, measured in grams per mole. Density is not measure in moles, so this is not the correct answer.

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c.The mass of substance dissolved in a liquid, measured in grams per milliliter. I think that this definition is correct but is incomple, so this answer is wrong.

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3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

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3 years ago

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tatyana61 [14]

$\\ \sf\longmapsto \Delta P=P$

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$\\ \sf\longmapsto m2=\dfrac{630}{10}$

$\\ \sf\longmapsto m2=63kg$

Bruces mass is 63kg

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2 years ago

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