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beks73 [17]
3 years ago
14

If there is an attractive force between all objects, why do we not feel ourselves gravitating toward massive buildings in our vi

cinity?
Physics
2 answers:
mars1129 [50]3 years ago
8 0
The mass of an object affects how powerful the attractive force is. To feel the pull / gravitation the mass of the object would have to be huge, bigger than that of massive buildings
HACTEHA [7]3 years ago
4 0
Because there is a far more massive nearby object that's attracting you to it. That object is the planet you're standing on. Compared to its mass, that massive building could just as well be a speck of dust or a clump of lint. The gravitational attraction toward Earth completely swamps out the attraction toward anything else.
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I’ll mark you as brinlist please help.
Ede4ka [16]

Answer:

245 divided by 5.14=47.6653696 or 47.66

Explanation:

8 0
3 years ago
Read 2 more answers
A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i
kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

\Delta A = 8.00-3.00= 5.00\ A

We need to calculate the time interval

Using formula of inductor

V=L\dfrac{\Delta A}{\Delta t}

\Delta t =\dfrac{L\Delta A}{V}

Where, \Delta A = change in current

V = voltage

L = inductance

Put the value into the formula

\Delta t=\dfrac{1.20\times5.00}{12.00}

\Delta t=0.5\ sec

Hence, The time is 0.5 sec.

5 0
3 years ago
Select the correct answer.
aleksklad [387]
A. Alternating current
4 0
2 years ago
How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
3 years ago
Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
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