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3 years ago

In what frame of reference would you be at rest while riding in a car?

Physics

2 answers:

Alexeev081 [22]3 years ago

8 0

In the frame of reference of anybody in the car.

gulaghasi [49]3 years ago

8 0

<h2>Answer:</h2>

<u>We would be in the frame of reference to other people inside the car.</u>

<h2>Explanation:</h2>

In Physics the frame of reference is used for the observation and mathematical description of physical phenomena that occur around us and they formulate the physical laws with respect to others. In the given case all the people inside the car are the frame of reference to one another but if someone is outside of the car then his frame of reference will be changing as the car moves with respect to him.

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Which process binds together sediment to form new rock?

vekshin1

Answer:

its cementation i got it correct

Explanation:

6 0

3 years ago

Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

kaheart [24]

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

$f'=f_{1}-f$ ...(I)

$f'=f_{1}+f$ ...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

$f'=462-5=457\ Hz$

$f'=462+5=467\ Hz$

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

4 0

2 years ago

In this dark-colored forest, you started with 50% light moths and 50% dark moths. At the end of your simulation, was there a hig

krek1111 [17]

Dark moths because they use their color to blend in with the trees to hide from birds

5 0

2 years ago

The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se

Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0

3 years ago

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

2 years ago