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3 years ago

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A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/

s.
a. Calculate the length of the incline
b. calculate the horizontal and vertical components of the velocity at the end of the incline.

1 answer:

exis [7]3 years ago

3 0

It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just

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What is the mass, in kilograms, of an avogadro's number of people, if the average mass of a person is 150 lb ?

Nikolay [14]

First step is to convert the lb to kg as follows:
1 lb = 0.45 kg
Therefore, 150 lb = 150 x 0.45 = 67.5 kg

Avogadro's number = 6.02 x 10^23

Mass of Avogadro's number of people = 6.02 x 10^23 x 67.5
= 4.0635 x 10^25 kg

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3 years ago

If the water in a lake is everywhere at rest, what is the pressure as a function of distance from the surface? The air above the

Rudiy27

Answer:

Answer:

101325 + 10055.25h

//

h = 10.1 m

Explanation:

the pressure at sea level = 1 atm = 101325 Pa

density of sea water = 1025 kg/ m^(3)

pressure due to fluid height = pgh

Absolute pressure = 101325 + 1025*9.81*h

= 101325 + 10055.25h

where h= 0 at sea level at increases downwards

//

101325 = 1025* 9.81* h

h = 10.1 m

Explanation:

7 0

3 years ago

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

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

$P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v$

Put the value into the formula

$P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}$

$P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}$

$P=0.350\ Watt$

Hence, The power output of the oscillator is 0.350 watt.

3 0

3 years ago

A 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same cons

LenKa [72]

Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
$F=ma$
from which we find an expression for the acceleration:
$a= \frac{F}{m}$ (1)

Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
$v_f^2 - v_i^2 = 2 a S$ (2)
where
$v_f$ is the final speed of the object
$v_i$ is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing $v_f$ (since the final velocity of the two objects is zero), we find
$-v_i^2 = 2 \frac{F}{m}S$
$S=- \frac{v_i^2 m}{2F}$
where we can ignore the negative sign (because the force F will bring another negative sign).

For the first object, we have
$S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m]$
And for the second object we have
$S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]$

And since the braking force applied to the two objects is the same, the two objects cover the same distance.

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4 years ago