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egoroff_w [7]
3 years ago
11

How are electric motors and generators similar? 

Physics
1 answer:
kotykmax [81]3 years ago
7 0

Answer:

they both produce energy

Explanation:

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A transverse, travelling wave on a chord is represented by = 0.22sin (5.6 + 34) where and are in meters and is in seconds. For t
Neko [114]

Answer:

A)  1.122 m

B)  5.4 hertz

C)  6.06 m/s ( towards negative  x axis )

D)  0.22

E)  max speed = amplitude * 34

                  = 0.22 * 34 = 7.48

  minimum speed ( speed at rest ) = 0

Explanation:

The travelling wave = D = 0.22sin ( 5.6x + 34t)

this wave is represented in the form : D = Asin ( kx + wt)

A) wavelength

k = 2\pi / v

v = wavelength

k = 5.6

therefore wavelength ( v ) from the equation = 2\pi / 5.6 = 1.122 m

B ) frequency

w = 2\pif

w = 34

f = frequency

therefore f = w / 2\pi  = 34 / 2\pi = 5.4 hertz

C ) velocity

speed = f v

v = 5.4 * 1.122 = 6.06 m/s ( towards negative  x axis )

D) amplitude

A = 0.22

E ) maximum and minimum speeds of particles on the chord

max speed = amplitude * 34

                  = 0.22 * 34 = 7.48

minimum speed ( speed at rest ) = 0

5 0
3 years ago
Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while
Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is t=2.7\ s

Malda initial velocity u=95\ cm/s

Let h be the height of Cliff

using h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

h=0+\frac{1}{2}(9.8)(2.7)^2

h=35.72\ m

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

R=u\times t

R=0.95\times 2.7=2.43\ m    

7 0
3 years ago
While driving down his street one evening, Jonah notices that his neighbor has laid out an electric fan for the garbage to pick
Kryger [21]

Answer: I have no idea either i need help aswell

Explanation:

4 0
3 years ago
Read 2 more answers
A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1
const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0
4 years ago
I need help ASAP doing a test rn
Irina-Kira [14]

Answer:

Thinking also friction force

Explanation:

7 0
3 years ago
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