ELECTROMAGNETIC FIELDS PLAN AN INVESTIGATION LAB REPORT FOR EDG.!!!!! WORTH 100 POINTS????????

A 0.42 kg shuffleboard disk is initially at rest when a player uses a cue to increase its speed to 4.2 m/s at constant accelerat

MariettaO [177]

Answer

given,

mass of the shuffleboard disk = 0.42 kg

speed of the cue is increased to = 4.2 m/s

acceleration takes over 2 m then acceleration is zero.

the disk additionally slide to 12 m

final speed of disk = 0 m/s

a) increase in thermal energy

$\Delta E_t = \dfrac{1}{2}m(v_1^2-v_2^2)$

$\Delta E_t = \dfrac{1}{2}\times 0.42 \times (4.2^2-v_2^2)$

$\Delta E_t = 3.704\ J$

b) $\Delta E_t = F_f.d$

F_f is the frictional force

$3.704= 12.F$

$F_f = 0.308\ N$

increase in thermal energy for entire movement of 14 m

$\Delta E_t = 0.308\times 14$

$\Delta E_t =4.312 \ J$

c) Work done on the disk by the cue

$W = \Delta KE + \Delta E_{t}$

$W = \dfrac{1}{2}\times 0.42 \times (4.2^2-0^2)+ F_f \times d$

$W = 3.704+ 0.308 \times 2$

W = 3.704 + 0.616

W = 4.32 J

3 0

4 years ago

2. A string with a length of 0.9m that is fixed at both ends

aliina [53]

Answer:

2a.) Wavelength = 1.8 m

2b.) F = 66.67 Hz

3a.) Find the attached file

3b.) Wavelength = 0.6 m

Explanation:

Given that the

Length L = 0.9m

Wavelength (λ) = 2L/n

Where n = number of harmonic

If n = 1, then

Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m

b.)

If waves travel at a speed of 120m/s on this string, what is the frequency

associated with the longest wave (first harmonic)?

Given that V = 120 m/s

V = Fλ

But λ = 2L, therefore,

F = V/2L

F = 120/1.8

F = 66.67 Hz

3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic

Using the same formula,

Wavelength (λ) = 2L/n

Where n = 3

Wavelength (λ) = 2 × 0.9/3

Wavelength (λ) = 0.6 m

3 0

3 years ago

Driving along a crowded freeway, you notice that it takes a time tt to go from one mile marker to the next. When you increase yo

ddd [48]

Answer:

38.6 mi/h

Explanation:

7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s

Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is

t = 1/v

Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means

t - 15 = 1/(v+0.00206)

We can substitute t = 1/v to solve for v

$\frac{1}{v} - 15 = \frac{1}{v + 0.00206}$

We can multiply both sides of the equation with v(v+0.00206)

v+0.00206 - 15v(v+0.00206) = v

$-15v^2 - 0.0308v + 0.00206 = 0$

$v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}$

$v= \frac{0.03083\pm0.35}{-30}$

v = -0.01278 or v = 0.01 0724 mi/s

Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h

6 0

4 years ago

Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be

miv72 [106K]

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

4 0

3 years ago

A musical note has a frequency of 512 Hz. If the wavelength of the note is 0.685 m, what is the speed of the sound of that note?

riadik2000 [5.3K]

Answer:

350.72 m/s

Explanation:

Formula for velocity of wave is;

v = fλ

Where;

v is speed

f is frequency

λ is wavelength

We are given;

f = 512 Hz

λ = 0.685 m

Thus;

v = 512 × 0.685

v = 350.72 m/s

5 0

3 years ago