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4 years ago

Is it true or false that on the Celsius temperature scale, there are no negative numbers? If false, why?

1 answer:

7 0

The Celsius Temperature scale has negative values, anything lower than 32 Fahrenheit/0° is negative, for reference using the celsius scale water freezes at 0° and boils at 100°, hope this helps!

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Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

OLga [1]

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

The position of the two cars at time t is given by the following functions:

$x_A(t) = \alpha t + \beta t^2$

with

$\alpha = 2.60 m/s\\\beta = 1.20 m/s^2$

Substituting,

$x_A(t)=2.60t+1.20 t^2$

And

$x_B(t)=\gamma t^2 - \delta t^3$

with

$\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3$

Substituting,

$x_B(t)=2.80t^2-0.20t^3$

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

$x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m$

$x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m$

So, car A is initially ahead.

The two cars are at the same point when their position is the same. Therefore, when

$x_A(t)=x_B(t)$

which means when

$2.60t+1.20t^2 = 2.80t^2-0.20t^3$

Re-arranging the equation, we find

$0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0$

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

$0.20t^2-1.6t+2.60=0$

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the same point.

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of car A:

$v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t$

The velocity of car B is given by the derivative of the position of car B:

$v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2$

Therefore, the distance between the two cars is not changing when the two velocities are equal:

$v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0$

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

The acceleration of each car is given by the derivative of the velocity of the car A.

The acceleration of car A is:

$a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40$

While the acceleration of car B is:

$a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t$

So, the two cars have same acceleration when

$a_A(t)=a_B(t)$

And solving the equation, we find:

$2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s$

So, the two cars have same acceleration at t = 2.67 s.

Learn more about accelerated motion:

brainly.com/question/9527152

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brainly.com/question/2506873

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#LearnwithBrainly

3 0

3 years ago

What are the differences between forces and pressure

slamgirl [31]

Answer:

Force is a push or pull action between objects. Pressure is force acting on a surface area of an object, thus it is force per unit area.

Explanation:

4 0

3 years ago

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

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7 0

2 years ago

N in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m. The den

Doss [256]

Answer:

The pressure is $P = 1.31*10^{5} \ Pa$

Explanation:

From the question we are told that

The depth of the swimming pool is $d = 3.00 \ m$

The density of water is $\rho = 1.00*10^{3} \ kg /m^3$

Generally the total pressure exerted on the bottom of the swimming pool is mathematically represented as

$P = P_o + \rho * g * h$

Here $P_o$ is the atmospheric pressure with value

$P_o = 101325 \ Pa$

$P = 101325 + [1000 * 9.8 * 3]$

=> $P = 130725 \ Pa$

=> $P = 1.31*10^{5} \ Pa$

7 0

3 years ago

Which form of energy is associated with an object’s position?

dybincka [34]

Answer:

potential

Explanation:

4 0

3 years ago

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