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shtirl [24]
3 years ago
11

Convert the following measurements as indicated, show work. write the answer in scientific notation.

Physics
2 answers:
klemol [59]3 years ago
7 0

Answer:

C. 17,000 m to kilometers

Explanation:

#CarryOnLearning

mojhsa [17]3 years ago
4 0

Answer:

A. 0.95 m

B. 1,100 ml

C. 17 km

D. 500,000 g

Explanation:

A. 95/100

B. 1.1 x 1,000

C. 17,000/1,000

D. 500 x 1,000

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A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0
3 years ago
A slender rod of length 80.0 cm and 0.600 kg has its center of gravity at its geometrical center. But its density is not uniform
Vlad [161]

Answer:

Icm = 0.0701 kgm^2

Explanation:

7 0
2 years ago
A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the acceleration the car presents during this time?
Art [367]

Answer:

1.02 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 m/s

Final velocity (v) = 6.6 m/s

Time (t) = 6.5 s

Acceleration (a) =.?

Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 6.6 m/s

Time (t) = 6.5 s

Acceleration (a) =.?

a = (v – u) / t

a = (6.6 – 0) / 6.5

a = 6.6 / 6.5

a = 1.02 m/s²

Therefore, the acceleration of the car is 1.02 m/s²

3 0
2 years ago
An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
2 years ago
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