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BartSMP [9]
3 years ago
7

Which of the following would probably not work with circuits on a daily basis

Physics
1 answer:
Anni [7]3 years ago
8 0
You need to attach the answer choices
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PLSSSSS HELP!!!!!!! ASAP!!!!! MARK BRAINLIEST!!!!
Ymorist [56]

Answer: I would say the answer is B.

Explanation: It looks and sounds the most correct.

5 0
3 years ago
The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in
kumpel [21]

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

5 0
3 years ago
Read 2 more answers
A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t
vladimir2022 [97]

Answer

Radius of the wheel r = 2.1 m

Moment of inertia I = 2500 Kg m²

Tangential force applied F = 18 N

Time interval t = 16 s

Initial angular speed ω1 = 0

Final angular speed ω2 = ?

Let α be the angular acceleration.

Torque applied τ = Iα

                         F r = Iα

Angular acceleration α = F r/I

                                    = \dfrac{18\times 2.1}{2500}

                                    = 0.015 rad/s²

(a)From rotational kinematic relation

            Final angular speed ω₂ = ω₁ + αt

                                                 = 0 + (0.015 rad/s^2 * 16 s)

                                                 = 0.24 rad/s

(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²

                      = 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0

                      =  72 J

(c) Average power supplied by the child P = W/t = \dfrac{72}{16}

                                                                        = 4.5 watt        

8 0
3 years ago
What is the weight of a 7.0 kilogram bowling ball on the surface of the moon
Artist 52 [7]
The acceleration on surface of moon =1.67m/s^2

Weight =mass ×acceleration
=7×1.67
=11.69N
8 0
3 years ago
An object moves at 60 m/s in the +x direction. As it passes through the origin it gets a 4.5 m/s^2 acceleration in the -x direct
Brut [27]

Answer:

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Explanation:

a) Let the starting position be origin and time be t.

  After time t displacement, s = 0 m

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

 We have equation of motion s = ut + 0.5 at²

 Substituting

        s = ut + 0.5 at²

        0 = 60 x t + 0.5 x (-4.5) x t²        

        2.25t² - 60 t = 0

        t² - 26.67 t = 0

        t (t-26.67) = 0

      t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

  Time , t = 26.67

 Substituting

        v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction

 

4 0
4 years ago
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