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3 years ago

Which state has the most fixed shape? O A. Gas O B. Solid O C. Liquid O D. Plasma

Physics

1 answer:

Mrac [35]3 years ago

3 0

Answer: Liquid

“A substance will take on the shape of an open container if it is a Liquid. Explanation: The major state of matter are solid, liquid and gas. Liquid usually have a definite volume.”

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An aluminum wire having a cross-sectional area equal to 3.00 10-6 m2 carries a current of 7.00 A. The density of aluminum is 2.7

Natali [406]

Answer:

Vd = 2.42 ×10⁻⁴ m/s

Explanation:

Given: A = 3.00×10⁻⁶ m², I = 7.00 A, ρ = 2.70 g/cm³

To find Drift Velocity Vd=?

Sol

the formula is Vd = I/nqA (n is the number of charge per unit volume)

n = No. of electron in a mole ( Avogadro's No.) / Volume

Volume = Molar mass / density ( molar mass of Al =27 g)

V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³

n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)

n= 6.02 × 10 ²⁸

Now

Vd = (7A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.00×10⁻⁶ m²)

Vd = 2.42 ×10⁻⁴ m/s

6 0

3 years ago

Can someone help label these?

seropon [69]

A. reactants
B. subscript
C. coefficient
D. products

7 0

2 years ago

a particle of mass 1.3kg is sliding down a frictionless slope inclined at 30 to the horizontal. the acceleration of the particle

OverLord2011 [107]

Answer:

$4.9\ m/sec^2$

Explanation:

The computation of acceleration of the particle down the slope is shown below:-

data provided in the question

Particle of mass = 1.3 kg i,e sliding down

Inclined = 30 to the horizontal

based on the above information

Force is given by

$N = mg\ cos \theta$ ............ 1

and sliding force is given by

$F = mg\ sin\alpha$

$a = g(sin\ 30^{\circ})$

$= 9.8\times \frac{1}{2} m/sec^2$

= $= 4.9\ m/sec^2$

Hence, the acceleration of the particle down the slope is 4.9 m/sec^2

4 0

3 years ago

An electric current produces a magnetic field.<br> a. True<br> b. False

Agata [3.3K]

The answer is true, an electric current does produce a magnetic field.

3 0

3 years ago

Read 2 more answers

The equation for free fall at the surface of a celestial body in outer space (s in meters, t in seconds) is sequals10.04tsqua

Sindrei [870]

Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

$s=\frac{1}{2}at^2$

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

$s=10.04 t^2$

this means that

$\frac{1}{2}g = 10.04$

so the acceleration of gravity on the body is

$g=2\cdot 10.04 = 20.08 m/s^2$

The velocity of an object in free fall starting from rest is given by

$v=gt$

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

$t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s$

3 0

3 years ago