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3 years ago

A certain part of a flat screen TV has a thickness of 150 nanometers. How many meters is this?

Physics

1 answer:

Bess [88]3 years ago

5 0

Answer:

1.5e-7 meters

.00000015 meters

Explanation:

.000000001 meters = 1 nanometer. Multiply that by 150 and an answer is there.

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An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0

3 years ago

When using the max out method to determine muscular strength, you must first do a warm-up set and a complete set before attempti

dlinn [17]

Answer:

True

Explanation:

I took the quiz .

5 0

3 years ago

Read 2 more answers

A small object with mass 1.30 kg is mounted on one end of arod

Julli [10]

Answer:

(a) $I_{system} = 1.014\ kg.m^{2}$

(b) $\tau = 0.0179\ N-m$

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, $\omega = 5010\ rev/min$

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

$I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}$

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = $2.30\times 10^{- 2}\ N$

$\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m$

3 0

3 years ago

What happens to the temperature of a substance during a phase change?

german

During a phase change, the temperature remains constant. obviously all the heat is used in phase transformation.

8 0

3 years ago

A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r

KIM [24]

Answer:

$\omega_f = 1.08 rad/s$

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

$L_i = L_f$

$I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f$

here we know that

$I_o = \frac{1}{2}mR^2$

$I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2$

$m_1 = 9.41 kg$

$r_1 = \frac{4}{5}(1.61) = 1.29 m$

$m_2 = 21.1 kg$

Now from above equation

$118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f$

$118.1(1.73) = 188.45\omega_f$

$\omega_f = 1.08 rad/s$

8 0

3 years ago