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damaskus [11]
3 years ago
15

A person who weighs 670 N steps onto a spring scale in the bath- room, and the spring compresses by 0.79 cm. (a) What is the spr

ing con- stant
Physics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

The spring constant is 84,810.13 N/m

Explanation:

Given;

applied force on the spring scale, f = 670 N

extention of the spring, x = 0.79 cm = 0.0079 m

Apply Hook's law to determine the spring constant;

f = kx

where;

f is the applied force

k is the spring constant

x is the is the extension

k = f / x

k = 670 / 0.0079

k = 84,810.13 N/m

Therefore, the spring constant is 84,810.13 N/m

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UN BLOQUE TIENE DE DIMENSIONES AB = 4 cm, BC = 2.5 cm, BF = 6 cm Y UNA MASA DE 300 gr  DETERMINE SU VOLUMEN  HALLE SU DENSIDAD
torisob [31]

Responder:

- Volumen del bloque = 60cm³

- Densidad = 5g / cm³

Explicación:

Dada la dimensión del bloque

AB = 4 cm, BC = 2.5 cm y BF = 6 cm

Volumen del bloque = Largo × Ancho × Alto

Volumen del bloque = AB × BC × BF

Volumen del bloque = 4 × 2.5 × 6

Volumen del bloque = 60cm³

Dada la masa del bloque = 300 gramos

Densidad = Masa / Volumen

Densidad = 300g / 60cm³

Densidad = 5g / cm³

Algunos materiales que tienen una densidad de 5g / cm³ o cercanos son radio, germanio y europio con densidades de 5g / cm³, 5.3g / cm³ y 5.24g / cm³ respectivamente.

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Defination of astronomy
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The branch of science which deals with celestial objects, space, and the physical universe as a whole.

Online definition so if you’re using it for school work I recommend re-wording!
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What technique is used by all three branches of science to make discoveries
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3 years ago
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

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If an ant were scaled up to the size of an elephant, the ant would be
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Answer:

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Explanation:

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