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Olin [163]
3 years ago
13

Select the community facility which typically has a variety of offerings for physical activity like team sports, exercise classe

s, and swimming.
Bowling Alley
Golf Course
Skate Park
YMCA
Physics
2 answers:
Yanka [14]3 years ago
5 0

Answer: It should be YMCA if im not wrong i took the test a while back.:)

Explanation:

ad-work [718]3 years ago
4 0

Answer:

YMCA

Explanation:

i know everything ;)

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Hair can be used to measure humidity.<br> a. True<br> b. False
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A. True.

You can make a hygrometer using strands of hair.

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3 years ago
A block with mass ma = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second
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7 0
4 years ago
An inductor is connected to an AC power supply having a maximum output voltage of 3.00 V at a frequency of 280 Hz. What inductan
Sergio [31]

Answer:

L = 0.48 H

Explanation:

let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax  be the maximum voltage and XL be the be the reactance of the inductor.

Vrms = Vmax/(√2)

         = (3.00)/(√2)

         = 2.121 V

then:

XL = Vrms/I  

     = (2.121)/(2.50×10^-3)

     = 848.528 V/A

that is L = XL/(2×π×f)

              = (848.528)/(2×π×(280))

              = 0.482 H

Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.

6 0
3 years ago
A device that has the capacity to receive and store electrical energy is a(n)
sergeinik [125]
The answer to your question is capacitor :)
6 0
3 years ago
Read 2 more answers
The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 5.00 T/s. Part A) What is the electric field str
Veronika [31]

Answer:

(A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is 3.75\times10^{-2}\ V/m.

Explanation:

Given that,

Magnetic field = 2.0 T

Diameter = 5.0 cm

Rate of decreasing in magnetic field = 5.00 T/s

(A). We need to calculate the electric field strength inside the solenoid at a point on the axis

Using formula of electric field inside the solenoid

E=\dfrac{r}{2}|\dfrac{dB}{dt}|

Electric field on the axis of the  solenoid

Here, r = 0

E=\dfrac{0}{2}\times5.00

E = 0

The electric field strength inside the solenoid at a point on the axis is zero.

(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis

Using formula of electric field inside the solenoid

E=\dfrac{r}{2}|\dfrac{dB}{dt}|

E=\dfrac{1.50\times10^{-2}}{2}\times|5.00|

E=0.0375= 3.75\times10^{-2}\ V/m

Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is 3.75\times10^{-2}\ V/m.

4 0
4 years ago
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