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Dahasolnce [82]

3 years ago

15

A block with a mass of 9.00 kg is pulled at a constant speed across a horizontal tabletop with a spring scale. The scale reads 6

1.8 N. Calculate the coefficient of kinetic friction

1 answer:

snow_tiger [21]3 years ago

7 0

Answer:0.69

Explanation:

Coefficient of kinetic friction=f/R=61.8/90=0.69

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Cual es el deporte que le da fortaleza y flexibilidad al cuerpo

AlexFokin [52]

Answer:

Aesthetic sports

Explanation:

Aesthetic sports are the one's that need well-developed physical qualities such as strength, agility, stamina, flexibility, and technical knowledge and artistry, in addition to technical ability and artistry. Elite athletes in these sports generally have a low abdominal fat , and the ranking is subjective.

In aesthetic sports like gymnastics, swimming, and figure skaters, dynamic and proactive flexibility is required.

4 0

3 years ago

A system consists of three particles with these masses and velocities: mass 3.00 kg, moving north at 3.00 m/s; mass 4.00 kg, mov

lions [1.4K]

Answer:

the total momentum is 8 .2 kg m/s in north direction.

Explanation:

given,

mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s

mass(m₂) 4.00 kg, moving south at v₂ = 3.70 m/s

mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s

north as the positive axis

south as the negative axis

now

total momentum = m₁v₁ + m₂ v₂ + m₃ v₃

total momentum = 3 x 3 - 4 x 3.7 + 7 x 2

= 9 - 14.8 + 14

= 8 .2 kg m/s

hence, the total momentum is 8 .2 kg m/s in north direction.

7 0

3 years ago

A 70-kg astronaut is space walking outside the space capsuleand is stationary when the tether line breaks. As a means of returni

Bogdan [553]

Answer:

0.4 m/s

Explanation:

Law of conservation of momentum tell us that the change in momentum of the hammer will be equal to the change in momentum of the astronaut

change in momentum of hammer = change in momentum of astronaut

2 kg (14 m/s - 0 m/s) = 70 kg * (v-0)

v = 0.4 m/s

7 0

3 years ago

Read 2 more answers

A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

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

$2.8 \text{ m} = 0+\dfrac{gt^2}{2}$

$t = \sqrt{\dfrac{5.6}{g}}$

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g </em> (It is going up)

Using the equation of motion

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

$1.2\text{ m} = ut-\frac{1}{2}gt^2$

Substituting the value of <em>t</em> from the previous equation,

$1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}$

$u\sqrt{\dfrac{5.6}{g}} = 4.0$

Taking <em>g</em> = 9.8 m/s²,

$u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}$

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

$u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}$

The initial velocity is

$v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}$

4 0

4 years ago

During the final stages of descent, asky diver with an

OLEGan [10]

Answer:

Gravity: downwards

Air drag and air-pressure on the inner surface of the the parachute: Upwards

Explanation:

If a sky-diver is in the final stages of his descend with open parachute such that the wind is calm and it does not blows him laterally.

In such a condition the air resistance in the form of drag and the pressure force due to the air captured in the parachute are acting in the upward direction which balance the force of gravity on the body. But this situation may occur momentarily and then again the diver must begin to slowly descend.

3 0

3 years ago