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Novosadov [1.4K]
3 years ago
14

There is a 12 v potential difference between the positive and negative ends of the jumper cables, which are a short distance apa

rt. an electron at the negative end ready to jump to the positive end has a certain amount of potential energy. on what quantities does this electrical potential energy depend? view available hint(s)
Physics
1 answer:
steposvetlana [31]3 years ago
4 0
The electric potential energy of the electron depends on the potential difference applied between the two ends of the cable. Indeed, the electric potential energy of a charge is given by
U=q \Delta V
where q is the magnitude of the charge, while \Delta V is the potential difference applied. So, U depends on \Delta V.
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The law of universal gravitation states that any two objects in the universe, without exception,
STatiana [176]
A. attract each other.

The Law of Universal Gravitation discusses the phenomenon of gravity. Remember that gravity is the force that keeps us on Earth; the Earth pulls us down, and our bodies pull back.  Gravity is the force of attraction, so the correct answer is a).
3 0
3 years ago
A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote
notka56 [123]

The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

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

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855#SPJ1

#SPJ1

8 0
1 year ago
A 0.60 kg rubber ball has a speed of 2.0 m/s at point A, and kinetic energy of 7.5 J at point
aliina [53]
<span>Let's first off calculate the kinetic energy using the formula 1/2MV^2. Where the mass, M, is 0.6Kg. And speed, V, is 2. Hence we have 1/2 * 0.6 * 2^2 = 1.2J. Since kinetic energy is energy due to motion; hence at point B the rubber has a KE of 1.2J and not 7.5J. So I would say that only the Mass and speed is actually true; While it's kinetic energy is not true.</span>
7 0
3 years ago
The fluid friction that opposes the motion of objects through air is known as what?
PSYCHO15rus [73]
The answer is Air Resistance
6 0
3 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
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