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2 years ago

Where is the fulcrum of a broom?

2 answers:

8 0

It's one of your hands. Which one it is depends on how you sweep.

-- If you hold the top of the stick motionless and wave your bottom hand
back and forth, then your top hand is the fulcrum, and you're using the
broom as a Class-3 lever.

-- If you hold your bottom hand motionless and wiggle the top end of the
broom back and forth with your top hand, then your lower hand is the fulcrum,
and you're using the broom as a Class-1 lever.

tatyana61 [14]2 years ago

5 0

It is the hand that holds the broomstick when you sweep

You might be interested in

What net force would be needed to accelerate the box at 2.3 m/s^2

Akimi4 [234]

The pertinent equation here is F=ma. You haven't shared the mass of the box, so I will use M to represent that mass.

Then F = M(<span>2.3 m/s^2) (answer)</span>

6 0

2 years ago

Read 2 more answers

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

$E_{m} = E_{k} + E_{p}$

$E_{m} = \frac{1}{2}v^{2} + gh$

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

$E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg$

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

$P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW$

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0

2 years ago

an object of mass m is traveling at constant speed v in a circular path of radius r. how much work is done by the centripetal fo

vlada-n [284]

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

Centripetal force - a force acts on an moving object in circular path.

the centripetal force is given by

F= mv²/r (equation1)

Work done is given by

W = Fd (equation 2)

d = 2π

work is done by the centripetal force on mass m during an angular displacement of 2π revolutions is given by:

to calculate work done using equation 1 in 2 we get

W = mv² d/r

W = mv² × 2π /r J

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

To know more about centripetal force :

brainly.com/question/13031430

#SPJ4

6 0

1 year ago

Someone plzzz helpppppp with this last question

vredina [299]

Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

7 0

2 years ago

During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consist

grandymaker [24]

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a) the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b) the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c) the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

6 0

2 years ago