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3 years ago

How is grounding a positive object different from grounding a negative object?

Physics

1 answer:

Natali5045456 [20]3 years ago

8 0

Answer:

Grounding a Positively Charged Object

Electrons were transferred from the electroscope to the ground. As in the case of grounding a negatively charged electroscope, the grounding of a positively charged electroscope involves charge sharing. The excess positive charge is shared between the electroscope and the ground.

Explanation:

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A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

valentina_108 [34]

F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer

7 0

3 years ago

Read 2 more answers

A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.

rodikova [14]

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

3 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago

Which group number below represents the Oxygen Group?

worty [1.4K]

Oxygen has an atomic number 8, because it has 8 protons and 8 electrons.

The first shell of an atom can hold up to 2 electrons but oxygen has 8 electrons, in that eight electrons 2 are in the first shell, so it has 6 more electrons left. The second shell can hold up to 8 electrons, oxygen has only 6 more electrons after the first shell is full, so it will have 6 electrons in the second shell

From this we know that oxygen has 2 shells so it is in period 2, and by counting from left to right, the sixth box in period 2 lies on group 16

Therefore Oxygen lies on group 16 and period 2

Happy to help :)

If you need my help for any other question, feel free to ask

8 0

3 years ago

A 51-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37.0Â° above the hor

fiasKO [112]

work is done by the pulling force which is same as the tension force in the rope. the net work done is zero for the crate since crate moves at constant velocity. but there is work done by the tension force which is equal in magnitude to the work done by the frictional force.

T = tension force in the rope = 115 N

d = displacement of the crate = 7.0 m

θ = angle between the direction of tension force and displacement = 37 deg

work done on the crate is given as

W = F d Cosθ

inserting the values given above

W = (115) (7.0) Cos37

W = 643 J

3 0

3 years ago