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Ad libitum [116K]
3 years ago
7

Is there is only one tool used to heat equipment in the laboratory.

Physics
2 answers:
yaroslaw [1]3 years ago
7 0
This is false there are many tools used to heat equipment in the laboratory such as a Fuel burner.
Dafna1 [17]3 years ago
6 0
No there is not only one tool used. They are many. Examples are:

Buchner funnel
Bunsen burner
Buret.
You might be interested in
If the man pushes with a force of 2000N and friction is 500N, what is the resultant force?
alexandr402 [8]

Answer:

Explanation:

If the force of 2000 N is directed towards the right and the friction is directed towards the left, the 2000 N force is positive and the other is negative. To find the resultant force:

2000 - 500 = 1500 N to the right

3 0
3 years ago
3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
3 years ago
A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring
AleksandrR [38]

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

v^2-u^2=2as

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

0-(17.7)^2=2a\times 33.1

a=\dfrac{-(17.7)^2}{33.1}

a=-9.46\ m/s^2

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

F = ma

F =1400\times(-9.46)

F=-13244\ N

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

|F| =13244\ N

Hence,  The magnitude of the horizontal net force is 13244 N.

5 0
3 years ago
Write one example of a physical change and one example chemical change
mojhsa [17]
Physical change = changes the physical properties (more commonly known as it's look)
Chemical change = changes the chemical properties into an entire new chemical form
Examples of physical change would be melting ice cubes or sugar cubes.
Examples of chemical change would be cooking eggs or burning paper because you're changing its chemical properties.
6 0
3 years ago
Which of the following statements is true about tides? (2 points)
andre [41]
1: <span>Tides are caused by the location of the Earth, sun, and moon.

2: </span><span>Tides occur about an hour later each day.
This is because of the earth's orbit and the moons position as they move.
</span><span>
3: </span>Half as much as the moon
The moon affects the tide much more than the sun, as the moon is drastically closer to the earth than the sun.
4 0
3 years ago
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