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3 years ago

Which force can affect an object without direct physical contact?

Physics

2 answers:

denis-greek [22]3 years ago

8 0

Gravity affet everything and it touches nothing.
Hope this helps!

kramer3 years ago

7 0

Gravity affets everthing it touches

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Water flows over the edge of a waterfall at a rate of 1.2 x 10^6 kg/s. There are 50.0 m between the top and bottom of the waterf

Anit [1.1K]

Answer:

5.88×10⁸ W

Explanation:

Power = change in energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

4 0

3 years ago

Read 2 more answers

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago

You're an electrical engineer designing an alternator (the generator that charges a car's battery). Mechanical engineers specify

Julli [10]

Answer:

13.78 mT

Explanation:

The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,

A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength

So,

B = ε/ωNA

substituting the values of the variables into the equation given that ε = 17 V

So, B = ε/ωNA

B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)

B = 17 V/(1233.8634 rad-turns-m²/s)

B = 0.01378 T

B = 13.78 mT

8 0

2 years ago

when a constant force acts upon an object the acceleration of the object varies inversely with its mass. when a certain constant

solong [7]

Explanation:

When a constant force acts upon an object the acceleration of the object varies inversely with its mass.

$a\propto \dfrac{1}{m}$

$\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}$

If m₁ = 21 kg, a₁ = 3 m/s², m₂ = 9 kg

We need to find a₂

So,

$a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{21\times 3}{9}\\\\a_2=7\ m/s^2$

So, if mass is 9 kg, its acceleration is 7 m/s².

8 0

3 years ago