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Anna71 [15]
2 years ago
8

In an experiment in which molten naphthalene is allowed to cool, the cooling curve shown below was obtained, the temperature 80∘

C is known as
A. cooling temperature

B. boiling point

C. melting point

D. vaporization point

Physics
1 answer:
notka56 [123]2 years ago
7 0

Answer:

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Explanation:

In an experiment in which molten naphthalene is allowed to cool, the cooling cur...

Question

In an experiment in which molten naphthalene is allowed to cool, the cooling curve in fig 5 was obtained, the temperature

80

∘

<h2>C.</h2>

is known as

Options

A) cooling temperature

B) boiling point

C) melting point √√√√√√√√√√※※※ ↑↑ANSWER

D) vaporization point

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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th
Igoryamba

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

5 0
3 years ago
What si unit measures speed ?
dlinn [17]
Hmm doesnt soujd familiar
4 0
3 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
A circuit is set up such that it has a current of 8 A. What would be the new current if the resistance was increased by a factor
RUDIKE [14]

Answer:

4 A

Explanation:

The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

V=RI

where

V is the voltage

R is the resistance

I is the current

The equation can also be rewritten as

I=\frac{V}{R}

from which we see that the current is inversely proportional to the resistance, R.

In this problem, the initial current is I = 8 A. Then the resistance is doubled:

R ' = 2R

So the new current is

I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}(\frac{V}{R})=\frac{I}{2}=4 A

so the current is halved.

7 0
3 years ago
Jjjjjjjjjjjjjjjjjjjjjjj<br>yyyyyyyyyyyyyyyy
MA_775_DIABLO [31]

Answer:

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Explanation:

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thank you

3 0
3 years ago
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