What is the formula for speed or velocity?

An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v

Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

$\ (x_{+ve}) = \frac{1}{2} \times 15 \times 20 \\\\$

$= \frac{1}{2} \times 300 \\\\= 150 \\\\$

Calculating the value of negative displacement upon the time t:

$(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\$

$= \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\$

$\to X= X_{+ve} + X_{-ve} \\\\$

$\to 150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25$

That's why its value lie in "between 20 s and 30 s".

6 0

3 years ago

two charges experience exert a force of 1n on each other when they are 1m apart. what force will these charges experience if the

Kisachek [45]

The force experienced by the charges when they are placed 2 m apart will be 0.25 N.

The force between two charges is given by Coulomb’s Law.

According to this law, the force between two charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges.

The above statement is represented by the following equation,

F= K Q1 x Q2/r^2

Here, F = Force between the two charges

K = Constant

Q1 and Q2 = Magnitude of the two charges

r = Separation between the two charges

According to this equation, the force is inversely proportional to the square of the distance between two charges.

Initially, when the two charges are 1 m apart, the force is 1 N.

When the distance between the two charges is doubled,

then according to Coulomb’s Law, the force should decrease by 4 times.

Hence, if the charges are placed 2 meters apart, the force becomes 1/4= 0.25 N.

To know more about "Coulomb's Law", refer to the following link:

brainly.com/question/9261306?referrer=searchResults

#SPJ4

8 0

1 year ago

If a seismic wave has a period of 0.0202s, find the frequency of the wave.

maw [93]

Answer:

49.5 Hz.

Explanation:

From the question given above, the following data were obtained:

Period (T) = 0.0202 s

Frequency (f) =?

The frequency and period of a wave are related according to the following equation:

Frequency (f) = 1 / period (T)

f = 1/T

With the above formula, we can obtain the frequency of the wave as follow:

Period (T) = 0.0202 s

Frequency (f) =?

f = 1/T

f = 1/0.0202

f = 49.5 Hz

Therefore the frequency of the wave is 49.5 Hz.

7 0

3 years ago

Raising the temperature of a gas will increase its pressure if the volume of the gas

ira [324]

That would be C .............

6 0

3 years ago

Which organisms develop gills from pharyngeal arches and later develop lungs to breathe on land

lawyer [7]

Apart from cutaneous respiration<span> present in all </span>species<span>, most lissamphibians are born in an aquatic larval stage with gills. After metamorphosis, they develop lungs to breathe on land. The larvae of urodeles and apods present external, filamentous and highly branched gills which allow them to breathe underwater.
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3 0

3 years ago

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