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3 years ago

8

Explian the construction of electric motor?

1 answer:

Nataly_w [17]3 years ago

7 0

Answer:

Hope the above picture might help you :)

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If you walk at an average speed of 5 km/h for 30 minutes, how

Inessa05 [86]

The distance that would be accumulated during the journey is 2.5 meters

The parameters given in the question are written below;

average speed= 5 km/hr

time = 30 minutes

convert 30 minutes to hours

= 30/60

= 0.5 hours

Distance-= speed × time

= 5 × 0.5

= 2.5 meters

Hence the distance of the entire journey is 2.5 meters

Please see the link below for more information

brainly.com/question/24268730?referrer=searchResults

3 0

2 years ago

A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth

yulyashka [42]

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi* $\sqrt{l/g}$

hence putting values we get ,

the solution is in picture ,

please

Brain-list it or support me at my U-Tube channel " ZK SOFT&GAMING " I will be thankful

5 0

3 years ago

Does anyone know this one? Thanks

Inessa [10]

Answer:

$3.844\,*\,10^5$

So a=3.844 and b=5

Explanation:

Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

$384,400=3.844 \,*\,100,000= 3.844 \,*\,10^5$

with a=3.844, and "5" as the exponent of ten (so b=5)

6 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

Describe a situation when you might travel at a high velocity, but with low acceleration

Alisiya [41]

Reading a book in your warm, comfy seat ... in Row-27 of a
passenger airliner cruising at 450 miles per hour.

5 0

3 years ago