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3 years ago

What is the mass of a uniform meter rule?

Physics

2 answers:

ruslelena [56]3 years ago

5 0

Answer:

A uniform meter rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm. ... When the balancing mass is moved then the resultant moment is the difference of clockwise moment and anticlockwise moment.

Reil [10]3 years ago

4 0

Answer:

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Matthew is looking at two different atoms. One has 8 protons and 10 neutrons. The other substance has 8 protons and 12 neutrons.

Lunna [17]

Answer:

They are the same element.

Explanation:

Atom >>>>> Proton >>>> Neutron

A >>>>>>>> 8 >>>>>>>>> 10

B >>>>>>>> 8 >>>>>>>>> 12

From the table above we can see that both atoms have the same proton number.

Therefore, they are the same element because they have the same proton number which means that they have the same atomic number. The element in this case is existing as an isotope in that the atoms have the same proton number but different neutron number.

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3 years ago

Magnesium hydroxide is a common ___?

omeli [17]

Answer:

Magnesium hydroxide is the inorganic compound with the chemical formula Mg(OH)2. Magnesium hydroxide is a common component of antacids, such as milk of magnesia, as well as laxatives

Explanation:

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3 years ago

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AnnZ [28]

Answer:

the switching circuitry is opened the soft iron armature is replaced to tasty le macha the switching circuitry is closed explain what happens when the switch circuit is a wonder you know about the characteristics of open to close this which right take the example of an electric generator for an electric motor which rotates in half rotation and change the armature and just try it ok I am sorry

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3 years ago

A flat horizontal surface with an area of 518 cm^2 is inside a uniform electric field of 1.33 x 10^4 N/C. The angle between the

GarryVolchara [31]

Answer: 576.48 N*m^2/C

Explanation: In order to calculate the electric flux through the any surface we have to take into account the scalar product between the electric field vector and the normal vector to the surface.

So we have:

ФE= E*A= 1.33 * 10^4*0.0518* cos (33.2°)= 576.48 N*m^2/C

3 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago

What is the mass of a uniform meter rule?​

What is the mass of a uniform meter rule?