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2 years ago

What 2 layers have temperature increasing with elevation?

Physics

2 answers:

leonid [27]2 years ago

6 0

Answer:

Stratosphere and Thermosphere

Explanation: Both of these layers above earth heat as you go higher.

Ulleksa [173]2 years ago

4 0

Stratosphere and thermosphere

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Which of the following is not the result of nuclear decay.

svlad2 [7]

Answer:it’s c

Explanation:

7 0

3 years ago

Beaker A contains 100 mL of water at a temperature of 25 °C. Beaker B contains 100 mL of water at a temperature of 60 °C. Which

Novosadov [1.4K]

Answer:

Only option A is correct. Beaker A has lower kinetic energy than beaker B.

Explanation:

Step 1: Data given

Beaker 1 has a volume of 100 mL at 25 °C

Beaker B has a volume of 100 mL at 60 °C

Thermal energy = m*c*T

Thermal energy beaker A = 100 grams*4.184 * 25°C

Thermal energy beaker B = 100 grams *4.184*60°C

⇒ Since both beakers contain the same amount of water, the thermal energy depends on the temperature.

Since beaker B has a higher temperature, it has a higher thermal energy than beaker A

When we heat a substance, its temperature rises and causes an increase in the kinetic energy of its constituent molecules. Temperature is, in fact, a measure of the kinetic energy of molecules.

This means beaker B has a higher kinetic energy than beaker A

Potential energy doesn't depend on temperature. this means the potential energy of beaker A and beaker B is the same.

a. Beaker A has lower kinetic energy than beaker B. This is correct.

b. Beaker A has higher thermal energy than beaker B. This is false.

c. Beaker A has higher potential energy than beaker B. This is false.

d. Beaker A has lower potential energy than beaker B. This is false

e. Beaker A has higher kinetic energy than beaker B. This is false.

3 0

2 years ago

How to atoms behave in non-magnetic items?

Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

5 0

2 years ago

A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo

zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

Lets explain how to solve the problem

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero ( $u_{y}=0$ ).

The ball is thrown from a height 4 meters above the ground.

The height $h=u_{y}t+\frac{1}{2}gt^{2}$

Remember: the height is negative value because its below the point of

thrown (initial position)

h = -4 m , $u_{y}=0$ and g = -9.8 m/s²(downward)

Substitute these values in the rule above

⇒ $4=0-\frac{1}{2}(9.8)t^{2}$

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ t = 0.9035

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range $R=u_{x}*t$ , where $u_{x}$ is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ $30=u_{x}(0.9035)$ (divide both sides by 0.9035)

⇒ $u_{x}=33.20$ m/s

The pitching speed of your friend is 33.20 m/s 

4 0

3 years ago

The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle

kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0

3 years ago