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3 years ago

5

NEED ASAP!! A box of mass 10 kg requires 20 N to slide it across a surface. What is the weight of the box? What is the coefficie

nt of friction between the box and surface ?

1 answer:

kaheart [24]3 years ago

3 0

Answer: 7

Explanation: 7 is the superior number

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A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a

marshall27 [118]

Explanation:

The given data is as follows.

mass (m) = 0.160 kg, spring constant (k) = 8 n/m,

Maximum speed ( $v_{m}$ ) = 0.350 m/s

Formula for angular frequency is as follows.

$\omega = \sqrt{\frac{{k}{m}}$

$\omega = \sqrt{\frac{{8}{0.160}}$

$\omega$ = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

$\nu_{max} = A \omega$

A = $\frac{\nu}{\omega}$

= $\frac{0.35}{7.07}$

= 0.05 m

Hence, value of amplitude is 0.05 m.

(b) Displacement = 0.030 m

Formula for mechanical energy is as follows.

M.E = $\frac{1}{2}kA^{2}$

Putting the values into the above formula as follows.

M.E = $\frac{1}{2}kA^{2}$

= $\frac{1}{2} \times 8 \times (0.05)^{2}$

= $9.8 \times 10^{-3}$ Joule

For x = 0.03,

As, P.E = $\frac{1}{2} \times kx^{2}$

= $\frac{1}{2} \times 8 \times (0.03)$

= $3.6 \times 10^{-3}$

Hence, calculate the kinetic energy as follows.

K.E = M.E - P.E

= ( $9.8 \times 10^{-3}$ - $3.6 \times 10^{-3}$ ) J

= $6.2 \times 10^{-3}$ J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is $6.2 \times 10^{-3}$ J.

7 0

3 years ago

The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin wi

svlad2 [7]

Answer

given,

largest diameter of balls = 9.52 mm = 0.00476 m

radius = 0.00476

smallest diameter of ball = 2.38 mm = 0.00238 m

radius = 0.00119

viscosity = 1.5 Pa.s

density of the ball = 1.42 g/cm

$F = 6\pi \eta r V_t$

$F = \dfrac{mv}{t}$

$F = \dfrac{\dfrac{4}{3}\pi\ r^3\times (\rho-\sigma) \times 0.99 V_T}{t}$

$6\pi \eta r V_t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{t}$

$t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{6\pi \eta r V_t}$

$t= \dfrac{0.22 r^2 (\rho-\sigma) }{\eta}$

for small balls

$t= \dfrac{0.22\times 0.00119^2 (1460-1300)}{1.5}$

t = 0.033 ms

for larger ball

$t= \dfrac{0.22\times 0.00476^2 (1460-1300)}{1.5}$

t = 0.531 ms

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4 years ago

A rock resting high in a cliff is an example of an object with what type of energy?

Basile [38]

<span>potential energy! Because the rock has the ability to possible move! So :)</span>

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4 years ago

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oksano4ka [1.4K]

I think the answer would be A.

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3 years ago

Activity: Brigada Tahanan, "Pamilyang namamahalan, Sa paglilinis sama-sama"

Dimas [21]

Answer:

WOW YOU SPEAK BEAUTIFUL MANDARIN

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3 years ago