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Oksana_A [137]
3 years ago
15

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

 COMPARE LA DENSIDAD HALLADA, CON LAS DENSIDADES DE SOLIDOS QUE APARECEN EN LAS TABLAS Y DIGA A QUE MATERIAL CORRESPONDE O NO
Physics
1 answer:
torisob [31]3 years ago
6 0

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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Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t    where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec =  1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r  in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

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its d

Explanation:

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3 years ago
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An object with a mass of 6.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the amount of the force? R
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You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
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<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

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4 years ago
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