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worty [1.4K]
3 years ago
12

If 3.5 paper clips = 1.0 pencils and your paper is 1.5 pencils long, how many paper clips long is your paper?

Physics
1 answer:
Andrews [41]3 years ago
6 0
\bf \begin{array}{ccll}
paperclips&pencils\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
3.5&1.0\\
p&1.5
\end{array}\implies \cfrac{3.5}{p}=\cfrac{1.0}{1.5}\implies \cfrac{3.5\cdot 1.5}{1}=p
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Waves travel as "packets" of several waves. in these "packets," a wave travels at __________.
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A. the same speed as the wave energy 
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2 years ago
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Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0
3 years ago
Why is force not on a scalar quantity??
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you might know that force,
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3 0
3 years ago
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A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

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6 0
2 years ago
A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring
AleksandrR [38]

Answer:

The magnitude of the horizontal net force is 13244 N.

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Using equation of motion

v^2-u^2=2as

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

0-(17.7)^2=2a\times 33.1

a=\dfrac{-(17.7)^2}{33.1}

a=-9.46\ m/s^2

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

F = ma

F =1400\times(-9.46)

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Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

|F| =13244\ N

Hence,  The magnitude of the horizontal net force is 13244 N.

5 0
3 years ago
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