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3 years ago

In what way are gravitational and electrical forces similar?

Physics

1 answer:

Nadusha1986 [10]3 years ago

6 0

Answer:

D. Both occur between objects independently whether they are in contact or not.

Explanation:

- The gravitational force is a force that is exerted between two (or more) objects having mass. This force is always attractive and its magnitude is given by

$F=G\frac{m_1 m_2}{r^2}$

where G is the gravitational constant, m1 and m2 are the two masses, and r is the distance between the two masses.

- The electrical force is a force that is exerted between two (or more) objects having electrical charge. It can be either attractive or repulsive, depending on the sign of the two charges, and its magnitude is given by

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 are the two charges, and r the distance between the two charges.

Looking at both formulas, we see that the two forces are present even when the two objects are not in contact with each other (in fact, r can assume any value in the formula). They are said to be non-contact forces. Therefore, the correct option is

D. Both occur between objects independently whether they are in contact or not.

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What quantity does the Hamiltonian correspond to? Rest mass Potential energy Total energy Kinetic energy Bohr frequency Lagrangi

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Answer:

The correct option is: Total energy

Explanation:

The Hamiltonian operator, in quantum mechanics, is an operator that is associated with the<u> total energy of the system.</u> It is equal to the sum of the total kinetic energy and the potential energy of all the particles of the system.

The Hamiltonian operator was named after the Irish mathematician, William Rowan Hamiltonis denoted and is denoted by H.

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3 years ago

The following is current scientific evidence supporting the nebular theory on the formation of the solar system.

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A.the composition of the inner and outer planets, current observations of star formation, and the motion of the solar system I hope this helps

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4 years ago

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The velocity of a motor bike of mass 140 kg is increased to 40m/s from 20m/s in 2 seconds. Then the force required for this is?

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Answer:

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3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

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3 years ago

How many hydrogen atoms are there in the following compound: 4CHA (2 points)

pogonyaev

Answer:

answer a, 4

Explanation:

when the 4 is before the compound it applies to the whole compound

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3 years ago

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