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3 years ago

7

In one experiment the electric field is measured for points at distances r from a uniform line of charge that has charge per uni

t length λ and length l, where l≫r. In a second experiment the electric field is:

1 answer:

Yuki888 [10]3 years ago

8 0

Answer:

(a) A. Uniform line of charge and B. Uniformly charged sphere

(b) To three digits of precision:

λ = 1.50 * 10^-10 C/m

p = 2.81 * 10^-4 C/m^3

Explanation:

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Comparing sound waves with different frequencies and wavelengths traveling through air,

aleksklad [387]

Explanation:

The speed of sound wave only depends on the property of the medium like density and the bulk modulus of the medium particle. The speed of sound also depends on the temperature of the medium.

On comparing sound waves with different frequencies and wavelengths traveling through air, the speed of the wave doesn’t depend on the frequency or the wavelength. Hence, the correct option is (1).

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3 years ago

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) $C(4,4) = 1; C(15,4) = 1365$

$P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326$

b) $C(4,3) = 4; C(11,1) = 11$

$P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223$

c) $C(4,2) = 6; C(11,2) = 55$

$P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418$

d) $C(11,4) = 330$

$P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418$

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3 years ago

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$

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3 years ago

Please answer this question??

Varvara68 [4.7K]

ANSWER: Itna Bada answer Kisi Ko Pata chalega

EXPLANATION : please Manje brainliest karo man Jay he brainiest Karo .

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3 years ago

A 12 kg object speeds up from an initial velocity of 10 m:s-1

amid [387]

Momentum = m • v

Original momentum = m • 10 m/s north

Final momentum = m • 15 m/s north

Change = m • (15 - 10) m/s north

Change = m • +5 m/s north

Change = +60 kg-m/s north

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3 years ago