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3 years ago

Examples were friction is not useful

Physics

1 answer:

expeople1 [14]3 years ago

7 0

The old electricity meters with disk using sapphire bearings for friction to be minimal to not aggravate metering

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A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

$T=2\pi \sqrt{\frac{L}{g} }$

Where T is period

L is length of rod

g is acceleration due to gravity = $9.8m/s^{2}$

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this $L_{O}$

$L_{O} = 3/4 * 5.95m$

= 4.4625m

thus $T=2\pi \sqrt{\frac{L_{O} }{g} }$

$T=2\pi \sqrt{\frac{4.4625 }{9.8} }$

T= 4.24sec

8 0

3 years ago

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

3 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy

Vsevolod [243]

Answer:

$v = 4.2 \ m/s$

Explanation:

Given data:

Mass of the paper clip, $m = 1.5 \ g = 0.0015 \ kg$

Kinetic energy, $K = 0.013 \ \rm J$

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

$K = \frac{1}{2}mv^{2}$

$0.013 = 0.5 \times 0.0015 \times v^{2}$

$\Rightarrow \ v = 4.16 \ m/s$

$v = 4.2 \ m/s.$ (rounding to nearest tenth)

3 0

3 years ago

A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer

Mrac [35]

Explanation:

The given data is as follows.

$m_{1}$ = 57 kg, $m_{2}$ = 79 kg

$l_{1}$ = 6.5 m, $l_{2}$ = (6.5 - 1.9) m = 4.6 m

(a) The sum of torque ends about far end is as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1}$ = 0

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5$ = 0

T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b) Now, we will calculate the sum about close ends as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}$

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5$ = 0

T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.

8 0

3 years ago

Could anyone help with this? :)

bonufazy [111]

I think the answer might be b

4 0

3 years ago

Read 2 more answers