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2 years ago

3. During a tug-of-war, Team A pulls with a

Physics

2 answers:

Neko [114]2 years ago

8 0

Answer:

8000 - 5000 =3000

Explanation:

Damm [24]2 years ago

5 0

<u>Answer:</u> The net force on the rope is 3000 N towards Team B.

<u>Explanation:</u>

Balanced force is defined as the forces that gets cancelled out when equal forces are acting in the opposite direction. The net force on the system comes out to be 0.

Unbalanced force is defined as the forces that does not get cancelled out because unequal forces act on the system in opposite direction. The motion of the object is in the direction of net force.

According to the question:

Force applied by Team A on the rope in tug-of-war = 5000 N

Force applied by Team B on the rope in tug-of-war = 8000 N

Net force acting on the rope = 8000 - 5000 = 3000 N (acting towards Team B)

Hence, the net force on the rope is 3000 N towards Team B.

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2 years ago

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A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i

dlinn [17]

A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.

<h3>What is electric potential energy?</h3>

The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r

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Thus, the distance between the charges will be 3.34 x 10⁷ m.

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2 years ago

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

Lyrx [107]

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

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2 years ago