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3 years ago

If a force of 12 N is applied to an object

Physics

1 answer:

Varvara68 [4.7K]3 years ago

6 0

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{12}{4} \\$

We have the final answer as

Hope this helps you

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The decomposition of ammonia to the elements is a first-order reaction with a half-life of 200 s at a certain temperature. How l

Feliz [49]

Based on the half-life of the reaction, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

<h3>What is the half-life of a substance?</h3>

The half-life of a substance is the time taken for half the amount of atoms present in that substance to decay.

The half-life of the reaction is 200 seconds.

After, one half-life, pressure reduces to 0.0500 atm

After, two half-lives, pressure reduces to 0.0250 atm

After, three half-lives, pressure reduces to 0.01250 atm

After, four half-lives, pressure reduces to 0.00625 atm

Time taken = 4 * 200 = 800 seconds

Therefore, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

Learn more about half-life at: brainly.com/question/26689704

#SPJ1

7 0

2 years ago

A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe

Rasek [7]

Answer:

$56.5\ \text{s}$

$21.13\ \text{m/s}$

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}$

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is $12.5+39+5=56.5\ \text{s}$

Distance traveled

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}$

$s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}$

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2$

$s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}$

The total displacement of the car is $156.25+975+62.5=1193.75\ \text{m}$

Average velocity is given by

$\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}$

The average velocity of the car is $21.13\ \text{m/s}$ .

6 0

3 years ago

A series of evenly timed pulses create a wave that can be described as a _______<br><br> wave.

son4ous [18]

Answer:

Periodic

Explanation:

A periodic (or repetitive) wave has continuously repeating pattern with characteristics such as amplitude, wavelength and frequency. In a periodic wave, a series of pulses that are evenly timed would be created. In other words, the wave pattern in a periodic wave repeats at regular intervals.

An example of a periodic wave is the sound from the strings of a violin.`

8 0

3 years ago

A child with a weight of 110 N swings on a playground swing attached to 2.00 m long chains. What is the gravitational potential

Ierofanga [76]

Answer

given,

Weight of the child = 110 N

length of the swing,L = 2 m

now, calculating the potential energy when the string is horizontal

Potential energy = m g h

now, h = L (1 - cos θ) where θ is the angle made by the string with the vertical.

PE = m g L (1 - cos θ)

when rope is horizontal θ = 90°

PE = 110 x 2 (1 - cos 90°)

PE = 220 J

now, calculating potential energy when string made 25° with horizontal

PE = m g L (1 - cos θ)

when rope is horizontal θ = 25°

PE = 110 x 2 (1 - cos 25°)

PE = 20.61 J

5 0

4 years ago

The cooking time for a roast scales like the 2/3rds power of the mass. Based on scaling laws, how much longer does a 20-lb roast

fgiga [73]

Answer:

2.726472 s more or 1.5874 times more time is taken than 10-lb roast.

Explanation:

Given:

- The cooking time t is related the mass of food m by:

t = m^(2/3)

- Mass of roast 1 m_1 = 20 lb

- Mass of roast 2 m_2 = 10 lb

Find:

how much longer does a 20-lb roast take than a 10-lb roast?

Solution:

- Compute the times for individual roasts using the given relation:

t_1 = (20)^(2/3) = 7.36806 s

t_2 = (10)^(2/3) = 4.641588 s

- Now take a ration of t_1 to t_2, to see how many times more time is taken by massive roast:

t_1 / t_2 = (20 / 10)^(2/3)

- Compute: t_1 / t_2 = 2^(2/3) = 1.5874 s

- Hence, a 20-lb roast takes 1.5874 times more seconds than 10- lb roast.

t_2 - t_1 = 2.726472 s more

3 0

3 years ago