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Ymorist [56]
2 years ago
4

Matt forgot to put the fabric softener in the wash. As his socks tumbled in the dryer, they became charged. If a small piece of

lint with a charge of +1.25 E -19 C is attracted to the socks by a force of 3.0 E -9 N, what is the magnitude of the electric field at this location?
3.8 E10 N/C


4.2 E10 N/C


2.4 E10 N/C


1.5 E10 N/C
Physics
1 answer:
il63 [147K]2 years ago
8 0

Answer: 2.4*10‐²⁴ N/C

Explanation:

Electric field intensity is the force acted per unit charge on a given surface

E = F/Q

Data

F = 3.0*10‐¹⁹ N

Q = 1.25 * 10-¹⁹ C

E = 3.0 * 10-⁹ / 1.25*10-¹⁹

E = 2.4*10¹⁰ N/C

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cicadas produce a sound that has a frequency of 123 Hz. what is the wavelength of this sound in the air? the speed of sound in a
mojhsa [17]

Answer: 2.72 metres

Explanation:

Given that:

frequency of sound F = 123 Hz. wavelength of sound in the air = ?

speed of sound in air V = 334 m/s

Recall that wavelength is the distance covered by the wave after one complete cycle. It is measured in metres, and represented by the symbol λ.

So, apply V = F λ

λ = V /F

λ = 334m/s / 123Hz

λ = 2.72m

Thus, the wavelength of this sound in the air is 2.72 metres

4 0
2 years ago
Water in the lake behind hoover dam is 221 m deep. part a what is the gauge water pressure at the base of the dam?
iragen [17]
<span>Answer: Pressure is always density * gravity * depth P = 1000 kg/m^3 * 9.81 m/s^2 * 221 m P = 2168010 Pa</span>
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2 years ago
Read 2 more answers
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

8 0
2 years ago
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

7 0
2 years ago
A rabbit is trying to cross the street. Its velocity v as a function of time t is given in the graph below where
Amiraneli [1.4K]

Answer:

2.5

Explanation:

8 0
2 years ago
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