Do two bodies have to be in physical contact to exert a force upon one another?

1 answer:

3 0

The correct answer to the statement " Do two bodies have to be in physical contact to exert a force upon one another " is:

No, the gravitational force is a field force and does not require physical contact to exert

a force.

The correct option is a.

<h3>Why two bodies do not have to be in physical contact to exert a force upon one another as a result of gravitational force.</h3>

It has been practically proven that two bodies can exert a force upon each other even if there is no physical contact between them. This can as a result of gravity.

That being said, a magnetic attraction can also exert a force between two different bodies upon one another.

So therefore, it can be deduced from above that two different bodies do not have to be in physical contact before they exert a force upon one another.

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The electrical force on a 2-c charge is 60 n. the electric field where the charge is located is

Kazeer [188]

The electrical force acting on a charge q immersed in an electric field is equal to
$F=qE$
where
q is the charge
E is the strength of the electric field

In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
$E= \frac{F}{q}= \frac{60 N}{2 C}=30 N/C$

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1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

love history [14]

Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$