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aalyn [17]
3 years ago
6

A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25

m from the left end, what is the mass of the meter-stick? (Hint: meter-stick is 1 m long). a) 0.6 kg b) 2.0 kg c) 1.0 kg d) 3.0 kg
Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

c) 1.0 kg

Explanation:

The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support

According to law of moment

Sum of clockwise moment = sum of anti clockwise moments

Clockwise moment = M×0.25(mass of metre rule is M)

CW moment = 0.25M

Anti clockwise moment = 0.25×1

ACW moments = 0.25kgm

Equate;

0.25M = 0.25

M = 0.25/0.25

M = 1.0kg

Hence the mass of the metre rule is 1.0kg

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The windshield of a car has a total length of arm and blade of 9 ​inches, and rotates back and forth through an angle of 93degre
Sergio039 [100]

Answer:

The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

Explanation:

Given

Length of blade = 9 inches

Angle of rotation = 93°

We're to calculate the area of the portion of the windshield cleaned by the 7​-in wiper​ blade?

We'll solve this by using area of a sector.

Area of a sector = ½r²θ

where θ is in radians.

So, angle of rotation (93°) must first be converted to radians

Converting 93º to radians, we get 31π/60

The area of the region swept out by the wiper blade = (area of the sector where r = 9 and

θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).

We're making use of 9-7 because that region is outside the boundary of the 7in blade

So Area = ½*9²*31π/60 - ½*2²*31π/60

Area = ½*31π/60(9²-7²)

Area = 31π/120 * (81 - 49)

Area = 31π/120 * 32

Area = 992π/120

Area = 62.49151386765697 in²

Area ≈ 62.49 in²

Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

5 0
3 years ago
Read 2 more answers
When light reflects from a surface, there is a change in its
alexandr402 [8]
The answer is: none of the above.

Explanation:

When light reflects from a surface, the frequency, wavelength, and speed do not change. They remain the same.
6 0
3 years ago
Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object
zhannawk [14.2K]

Answer:

c)

Explanation:

A collision is said to be elastic when the total kinetic energy is the same after the collision. The speed of objects that are stuck together will always be less than the initial speed of the object that was in motion given that the other particle was at rest. It is because the kinetic energy of the system was due to the moving object. The objects have a greater overall mass when they are stuck. If the kinetic energy is the same and the mass increases, the velocity must decrease.

3 0
3 years ago
The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0
3 years ago
Can you please help me find the answer
Serggg [28]
Im sorry may you please retake the picture then i will answer
8 0
3 years ago
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