Place the gears in order from highest to lowest torque?

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

7 0

3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

Why does a skier wear polarized glasses instead of glasses that are used for reading?

Afina-wow [57]

A skier wears polarized glasses instead of glasses that are used for reading because polarized glasses decrease reflected glare compared to regular glasses. (D)

Hope this answers your question correctly.

8 0

3 years ago

Read 2 more answers

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. find the minimum en

shusha [124]

The energy carried by the incident light is
$E=hf$
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
$E=hf=(6.63 \cdot 10^{-34}Js)(5.64 \cdot 10^{14}s^{-1})=3.74 \cdot 10^{-19}J$

4 0

4 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago