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svlad2 [7]
3 years ago
14

An object of mass 10.0 kg is initially at rest. A 100 N force causes it to move horizontally througha distance of 6.00 m along a

frictionless surface. What is the change in the kinetic energy of thisobject?
Physics
2 answers:
KonstantinChe [14]3 years ago
3 0

Answer:

Δ K.E = 600 J

Explanation:

Given that:

the mass of the object = 10.0 kg

Force = 100 N

distance = 6.00 m

what is the change in kinetic energy of this object = Δ K.E

The expression for the change in the kinetic can be illustrated as follows:

Δ K.E = W

where W is the workdone on the object and is given by the formula :

W = F × d

So;

Δ K.E = F × d

Δ K.E = (100 N × 6.00 m)

Δ K.E = 600 N.m

Δ K.E = 600 J

Therefore, the change in the kinetic energy of this object = 600 J

levacccp [35]3 years ago
3 0

Answer:

The change in the kinetic energy is 600 J

Explanation:

According the work energy expression, the change of kinetic energy is equal to the work done, the equation is:

deltaE_{k} =W=Fs

Where

F = force = 100 N

s = distance = 6 m

Replacing:

deltaE_{k} =100*6=600J

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zhuklara [117]

Answer:

1) El diámetro es de aproximadamente 913,987 cm.

2) La fuerza del cilindro es 5576850 kgf

Explanation:

1) Los parámetros dados son;

El volumen del aire = 13,122 litros = 13122000 cm³

La presión de trabajo = 8.5 kgf / cm²

La longitud del cilindro = 20 cm.

Por lo tanto, tenemos;

El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²

r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo

∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf

La fuerza del cilindro = 5576850 kgf

3 0
3 years ago
A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m
harina [27]

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

6 0
3 years ago
What information about an element is most crucial for locating that element in the periodic table?
castortr0y [4]
The most crucial information would be its atomic number. 
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3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

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<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

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2 years ago
Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple
Finger [1]
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This demonstration is a tremendous boost for the work of both Kepler
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5 0
3 years ago
Read 2 more answers
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