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leva [86]
2 years ago
12

1. To describe the length of a classroom, a student should use what? A) centigrams B) centimeters C) kilometers D) meters 2. Whi

ch is a standard metric base unit, correctly matched with what it measures?
A) Kelvin: time B) second: time C) meter: volume D) gram: temperature A tectonic plate near Niihau, an island in Hawaii, grows at an average rate of 11 cm/year. How many centimeters will the plate grow in 2 years?
A) 2 cm B) 11 cm C) 22 cm D) 30 cm The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will the plate move in 10 years? A) less than 5 cm B) equal to 25 cm C) less than 25 cm D) greater than 25 cm
Physics
2 answers:
Gemiola [76]2 years ago
7 0
1).  To describe the length of a classroom, a student can use any unit
he feels like using.  Some units will produce ridiculous numbers, though. 
The most convenient numbers will result from using meters or centimeters. 
If the room is, say, 20 feet long, then that's about 0.0061 kilometer, or
about 610 centimeters, or about 6.1 meters.
Meters is probably best.

2).
A). Kelvin: time.       No.  Kelvin is a unit of temperature.
B). second: time    Yes.  Second is a unit of time.
C). meter: volume    No.  Meter is a unit of length or distance.
D). gram: temperature.  No.  Gram is a unit of mass.

3).  A plate near Hawaii grows about 11 cm per year.
So it grows 11 cm in the first year, and another 11 cm in the second year.
The total growth in 2 years is (11cm + 11cm) = 22 cm .

4).  A slower plate in the Arctic moves slightly less than 2.5 cm per year.
It moves ...
slightly less than 2.5 cm in the 1st year
slightly less than 2.5 cm in the 2nd year
slightly less than 2.5 cm in the 3rd year
slightly less than 2.5 cm in the 4th year
slightly less than 2.5 cm in the 5th year
slightly less than 2.5 cm in the 6th year
slightly less than 2.5 cm in the 7th year
slightly less than 2.5 cm in the 8th year
slightly less than 2.5 cm in the 9th year
and
slightly less than 2.5 cm in the 10th year
for a grand total of 
           (slightly less than 2.5 cm x 10) = less than 25 cm in 10 years.

sdas [7]2 years ago
7 0

1 d,   2 b,   3 c,   4 c that is the answer to them

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5 0
3 years ago
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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

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4 0
3 years ago
We only see objects because they absorb light. <br> True or False?
Alenkasestr [34]

Answer:

true

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i think it's true because I took a quiz on this

3 0
3 years ago
A 795-loop square armature coil with a side of 10. 5 cm rotates at 70. 0 rev/s in a uniform magnetic field of strength 0. 45 t.
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The rms voltage output of the generator is 1.94 × 10⁻ ⁵ V.

RMS is an acronym for root mean squared. An RMS value is more than just the "amount of AC power that causes the same heating impact as an analogous DC power" or something along those lines.

No. of loop = 795

Diameter of the coil = 10.5 cm

Radius of the coil = 5.25 cm

Magnetic Field, B = 0.45 T

Time, t = 70.0 rev/s

              V_{rms} =\frac{NwAB}{\sqrt{2} }

Where,

              N = No. of loop

              A = Area of the coil

              B = Magnetic Field

              V_{rms} = Voltage rms

Area of the coil = πr²

                        = 86.57 cm²

w = 2π/t

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   = 0.089

V_{rms} =\frac{795*0.089*86.57* 0.45}{\sqrt{2} }\\\\V_{rms} =\frac{2756.36}{\sqrt{2} }\\\\\\V_{rms} =\frac{2756.36}{1.414 }\\\\V_{rms} = 1.94 * 10^-^5 V

Therefore, the rms voltage output of the generator is 1.94 × 10⁻ ⁵ V.

Learn more about rms voltage here:

brainly.com/question/13156072

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8 0
10 months ago
A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured
Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}

Put the value into the formula

0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}

v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}

v_{3}=4.98\ m/s

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0
3 years ago
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