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3 years ago

11

What conclusions can you draw about an object that either has an OVERALL negative charger OR an OVERALL positive charge?

1 answer:

sveticcg [70]3 years ago

6 0

Answer:

The object is also positively charged because same or alike charges repel

Explanation:

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According to research, college graduates are more likely to engage in all of the following EXCEPT:

Ksivusya [100]

Strange question. I’d say B)

7 0

3 years ago

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A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box

forsale [732]

We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.

From N-Gcosα=0 we get:

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.

8 0

2 years ago

Please help with this, oh and bungee gum contains both the properties of rubber AND gum you see?

Sladkaya [172]

Can I still get 5 points bc u already figured it out

8 0

3 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

3 years ago

The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium

Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × $10^{37}$ Hertz

Given:

Energy = 378 $\frac{kJ}{mol}$

To find:

Minimum frequency of light required to ionize magnesium = ?

Formula used:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

Solution:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

738 × $10^{3}$ = 6.63 × $10^{-34}$ × v

v = 111.31 × $10^{37}$ Hertz

The minimum frequency required to ionize the photon is 111.31 × $10^{37}$ Hertz

7 0

2 years ago

Read 2 more answers