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3 years ago

6

ANALOGY: Mars bar: planet as Milky Way: __________

1 answer:

strojnjashka [21]3 years ago

4 0

Galaxy is the answer

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After an oil spill in the ocean, what happens to bacteria that break down the oil?

Tpy6a [65]

Answer:

A

Explanation:

7 0

3 years ago

A student is conducting an experiment to determine how far a ball will roll down a ramp based on the angle of incline. What are

icang [17]

Controls or controlled variables are variables that should remain unchanged in the experiment. These variables can have a direct effect on the dependent variable. Take note that controlled variables are NOT what is being tested but can alter results. Three possible controls for this experiment would mainly be features of the ball and the ramp.

Examples and possible answers to your question would be:

1. Mass of the ball
2. Size of the ball
3. The density of the ball
4. Type of ball
5. Length of the ramp
6. Type of ramp
7. The thickness of the ramp
<span>
Other controls that you can consider also is the environment. Wind can also affect the ball that is being rolled.</span>

6 0

3 years ago

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 61.0 cm mark on the track. The glider compl

Novosadov [1.4K]

Answer:

T = 2.82 seconds.

The frequency $\mathbf{f = 0.36 \ Hz}$

Amplitude A = 25.5 cm

The maximum speed of the glider is $\mathbf{v = 56.87 \ rad/s}$

Explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

$T = \frac{31}{11}$

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

$f = \frac{1}{T}$

$f = \frac{1}{2.82}$

$\mathbf{f = 0.36 \ Hz}$

The amplitude is determined by using the formula:

$A = \frac{d}{2}$

The limits that the spring makes the oscillations are from 10 cm to 61 cm.

The distance of the glider is, d = (61 - 10 )cm = 51 cm

Replacing 51 for d in the above equation

$A = \frac{51}{2}$

A = 25.5 cm

The maximum speed of the glider is:

$v = A \omega$

where ;

$\omega = \frac{2 \pi}{T}$

$\omega = \frac{2 \pi}{2.82}$

$\omega = 2.23 \ rad/s$

$v = A \omega$

$v = 25.5 *2.23$

$\mathbf{v = 56.87 \ rad/s}$

3 0

3 years ago

You drive on Interstate 10 from San Antonio to Houston, half the time at 54 km/h and the other half at 118 km/h. On the way back

Irina-Kira [14]

Answer:

Average speed: 86 km/h

Explanation:

Driving from San Antonio to Houston:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h} }{2}=86 \frac{km}{h} [\tex]

Driving way back:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h} }{2}=86 \frac{km}{h} [\tex]

As in both routes we have the same average speed, then the average speed for the whole trip is 86 km/h

7 0

3 years ago

Read 2 more answers

A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson

Andrews [41]

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

$f = \dfrac{nv}{2L}$

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

$f = \dfrac{3\times 345}{2\times 0.96}$

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

7 0

3 years ago