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3 years ago

According to the kenitic theory, what happens when ice melts

2 answers:

8 0

Answer:Heat is kinetic energy the higher the temperature of a substance, the faster and farther its molecules move. For example, as heat transfers into the ice, the ice molecules move faster and eventually the ice melts When those molecules slow down, their kinetic energy decreases

Hope this helps! :)

ohaa [14]3 years ago

4 0

Answer:

The molecules begin to speed up.

Explanation:

When heat is applied to an object, the molecules inside begin to move faster and faster. Starting very slow at a solid and up to insanely fast at a water vapor.

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You are in the lab and are given two rods set up so that the top rod is directly above the bottom one. The two straight rods 50-

kompoz [17]

Answer:

Explanation:

The magnetic force due to lower rod must be equal to weight of upper rod for equilibrium .

magnetic field due to lower rod on upper rod

= ( μ₀ / 4π ) x(2i / r ) , i is current , r is distance between rod

= 10⁻⁷ x 2 x 15 / 1.5 x 10⁻³

= 20 x 10⁻⁴ T

force on the upper rod

= B i L , B is magnetic field , i is current in second rod and L is its length

= 20 x 10⁻⁴ x 15 x .50

= 150 x 10⁻⁴ N

= .015 N

This force can balance a wire having weight equal to .015 N .

= .00153 kg

= 1.53 g .

wire should weigh 1.53 g .

3 0

4 years ago

Look at the circuit diagram. <br><br> What type of circuit is shown?

Nimfa-mama [501]

Answer:

Its a parallel connection

Explanation:

The bulbs aren't in one line but two different parts, which makes it a parallel connection instead of a series connection.

7 0

3 years ago

Read 2 more answers

A block (mass m1) lying on a frictionless inclined plane is connected to a mass (mass m2) by a massless cord passing over a pull

azamat

Answer:

a) System aceleration:

$a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}$

b) Direction of movement:

The block $m_1$ down the plane when the acceleration is negative. This occur when:

$m_2-m_1sin(\theta)$

The block $m_2$ up the plane when the acceleration is positive. This occur when:

$m_2-m_1sin(\theta)>0$

Explanation:

For the block $m_1$ the move direction is parallel (||) to the plane

$\sum F_{||}=m_1a=T-sin(\theta)mg$ (1)

For the block $m_2$ the move direction is vertical (y)

$\sum F_y=m_2a=m_2g-T$ (2)

Both blocks are connected with the same cable, therefore, the tension on the cable and the acceleration is the same for both.

Solving the equation 2 for T:

$T=m_2g-m_2a$ (3)

replacing (3) in the equation (1)

$m_2g-m_2a- m_1gsin(\theta)=m_1a$ (4)

solving (4) for a:

$a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}$

8 0

3 years ago

How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?(gravity is 9.8)

Mekhanik [1.2K]

First, calculate the initial velocity of the dog given with the vertical height and the acceleration due to gravity which is calculated through the equation,
2ad = Vo²
Substituting the known values,
2(9.8 m/s²)(1.2 m) = V₀²
V₀ = 4.85 m/s
The kinetic energy is solved through the equation,
KE = 0.5mv²
Substituting the known values to the latest equation,
KE = 0.5 (7.2 kg)(4.85 m/s)²
KE = 17.46 J
Thus, the kinetic energy is 17.46 J.

8 0

3 years ago

A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov

Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0

3 years ago

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