How do we calculate the "range" of the measurements?

Physics

1 answer:

Alika [10]3 years ago

8 0

Answer: I think its A

Explanation:

If not, what is the answer, beause I'm on the same quiz.

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Hey everyone!

Soloha48 [4]

8. In soft magnetic materials such as iron, what happens when an external magnetic field is removed? a. The domain alignment persists. b. The orientation of domains fluctuates. c. The material becomes a hard magnetic material. d. The orientation of domains changes, and the material returns to an unmagnetized state. 9. According to Lenz’s law, if the applied magnetic field changes, a. the induced field attempts to keep the total field strength constant. b. the induced field attempts to increase the total field strength. c. the induced field attempts to decrease the total field strength. d. the induced field attempts to oscillate about an equilibrium value. 10. The direction of the force on a current-carrying wire in an external magnetic field is a. perpendicular to the current only. b. perpendicular to the magnetic field only. c. perpendicular to the current and to the magnetic field. d. parallel to the current and to the magnetic field

6 0

3 years ago

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$