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slega [8]
3 years ago
9

A man stands on a scale and holds a heavy object in his hands. What happens to the scale reading if the man quickly lifts the ob

ject upward and then stops lifting it?
Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

Explanation:

When he accelerates the heavy object up , the reading increases because an extra downward normal force acts on it, then scale reading returns to the same reading as when standing stationary, and then decreases as although he is lifting the heavy object , the acceleration is decreasing ,so the extra upward normal force acts.

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How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0
3 years ago
What is the impulse needed to stop a 45-kg boy who is running at 6 m/s in 3 seconds?
Tju [1.3M]
Impulse = change of momentum
Impulse = 45 x 6 = 270 Ns
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Using the image provided, which moon phase is next?
ella [17]

Answer:

Waning Crescent

5 0
3 years ago
Read 2 more answers
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:
Westkost [7]

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

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If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
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