Can someone list two non-examples of volcano formations/eruptions?

NHT0015

poizon [28]

Answer:

(1) 1×10⁻⁴

Explanation:

From the question,

α = (ΔL/L)/(ΔT)............. Equation 1

Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.

Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C

Substitute these values into equation 1

α = 1×10⁻⁴/10

α = 1×10⁻⁵ °C⁻¹ .

β = (ΔA/A)/ΔT................... Equation 2

Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.

make ΔA/A the subject of the equation

ΔA/A = β×ΔT.......................... Equation 3

But,

β = 2α.......................... Equation 4

Substitute equation 4 into equation 3

ΔA/A = 2α×ΔT................ Equation 5

Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹

Substitute into equation 5

ΔA/A = ( 2)×(1×10⁻⁵)×(5)

ΔA/A = 10×10⁻⁵

ΔA/A = 1×10⁻⁴

Hence the right option is (1) 1×10⁻⁴

3 0

3 years ago

A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc

Lana71 [14]

1. - 12N

2.- 2 hours

3. - 0.5 m/s/s

7 0

3 years ago

Read 2 more answers

At what altitude h above the north pole is the weight of an object reduced to 67% of its earth-surface value? Assume a spherical

lara31 [8.8K]

Answer:

The answer to the question is

At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value

Explanation:

We make use of the gravitational formula as follows

F = G $\frac{m_{1} m_{2} }{R^{2} }$ where

m₁ = mass of the object

m₂ = mass of the earth

d = distance between the two objects and

G = gravitational constant

if at the altitude the weight is reduced to 67 % of its weight on earth then

with all other variables remaining constant, we have

67% F = G $\frac{m_{1} m_{2} }{R_{2} ^{2} }$ =0.67× G× $\frac{m_{1} m_{2} }{R_{1} ^{2} }$

cancelleing like ternss from both sides we have

1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185

or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m

Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value

6 0

3 years ago

The length is a fundamental quantity why

kupik [55]

There’s no picture and what’s the length?

5 0

3 years ago

A women who normally weighs 400 N stands on top of a ladder so high she is one full additional earth radius above the surface of

AveGali [126]

The woman weighs 100N now.

Explanation:

Given:

F is given as 400N

one full additional earth radius formed

To Find:

Weight of the woman=?

Solution:

We know that gravitational force is inversely proportional to the square of radius.

$f_g=\frac{1}{r^2}$

where

$f_g$ is the gravitational force

$r^2$ is the radius square

So when radius is doubled, the force becomes one fourth of the original.

Therefore $\frac{1}{4} \times400 = 100N$

3 0

3 years ago