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2 years ago

I'll give brainliest!

2 answers:

8 0

Answer: Earth's magnetic field has flipped its polarity many times over the ... Earth has settled in the last 20 million years into a pattern of a pole reversal about ... And while reversals have happened more frequently in "recent" years, when ... per year, as opposed to about 10 miles per year in the early 20th century.

Explanation:

Mama L [17]2 years ago

3 0

the answer to this problem is 15

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Imma straight male btw also wut are newtons law write a paragraph desricbing them

tekilochka [14]

Third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

8 0

2 years ago

Potassium loses electrons when it reacts with oxygen. Which statement is true of potassium in this reaction? A. It reduces. B. I

allochka39001 [22]

<h2>Answer:</h2>

[D] It oxidizes.

4K + O2 → 2K2O, potassium oxide is produce.

We have the equation, in which when potassium react with oxygen, as a product it formed Potassium oxide which means adding of oxygen (oxidation).

5 0

2 years ago

Part II # 1 A mass on a string of unknown length oscillates as a pendulum with a period of 4 sec. What is the period if: (Parts

Mrac [35]

Answer:

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

$T=2\pi\sqrt{\frac{L}{g} }$ where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

$T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }$

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

$T_1=2\pi\sqrt{\frac{2L}{g} }=\sqrt{2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{2}*4=5.7\ sec$

3) The string length is halved

Given that:

$T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }$

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

$T_1=2\pi\sqrt{\frac{L}{2g} }=\sqrt{1/2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{1/2}*4=2.8\ sec$

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.

6 0

2 years ago

A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of lengt

damaskus [11]

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure, 41.0 PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be 14.6959 P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂ ⇒ V₂ = P₁*V₁ / P₂

where

1 and 2 are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in) ⇒ V₁ = 50.4 in³

P₁ = 14.6959 P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959 P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒ V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A = (13.2985 in³) / (3.00 in²)

⇒ h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

4 0

2 years ago

A shield of material that reflects neutrons is placed around a radioactive sample. Which change in the nuclear reaction is most

Crazy boy [7]

Answer:

A chain reaction will be sustained in a sub-critical mass.

Explanation:

Hope this helps!

If not, I am sorry.

3 0

1 year ago