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klio [65]
3 years ago
11

A security guard walks at a steady pace traveling 170 m in one trip around the perimeter of a building.

Physics
1 answer:
WARRIOR [948]3 years ago
6 0
Speed= distance/time
Speed= 170/230
Speed=0.74 m/s
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MIDDLE SCHOOL: What are the three types of energy used when cleaning? (Please write them in order of occurrence)
tankabanditka [31]

Answer:

Chemical, mechanical, thermal i guess

7 0
3 years ago
An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Bond [772]

The y-component of the acceleration is 0.33 m/s^2

Explanation:

The y-component of the acceleration is given by:

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the  final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s

where

u = 2.25 m/s is the initial velocity

\theta_1 = 50.0^{\circ} is the initial  direction

v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s, where

v = 4.65 m/s is the final velocity

\theta_2 = 120.0^{\circ} is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2

Learn more about acceleration:

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3 0
3 years ago
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
After the slab is charged, and the electrophorus is placed on the slab, the student briefly touches the electrophorus, effective
inna [77]

Answer:

By induction method

Explanation:

Induction method involves charging an electrically neutral body by bringing it in contact with an electrically charged body.

For the electrophus, a charge opposite that on the slab is induced on the side in contact with the slab; driving the opposite charge (this will be the same as that on the slab) to the other end of the elctrophus. Touching the electrophus removes the charge opposite the charge induced on the electrophus by the charged slab either by drawing up charge from the earth or taking the charge to earth (depends on the charge. A negative charge is drawn to earth while a positive charge draws up electrons from the earth)

5 0
3 years ago
A singer raises his pitch one octave. what quality of the sound wave has changed?
Keith_Richards [23]
I think the amplitude changed 
3 0
3 years ago
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