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2 years ago

How do the molecules of cold water differ from the molecules of hot water?

Physics

2 answers:

zmey [24]2 years ago

7 0

The molecules of cold water have less energy and are moving slower than the one with hot water

soldier1979 [14.2K]2 years ago

4 0

The molecules of cold water move slower then the molecules of warm water.

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A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Sunny_sXe [5.5K]

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ a = -7.088 m/s² [Answer]

Negative sign denotes that the velocity is decreasing.

4 0

2 years ago

Read 2 more answers

4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strik

nikklg [1K]

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ = 222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

5 0

3 years ago

Read 2 more answers

A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio

Bumek [7]

Answer:

This is known as a Galilean transformation where

V' = V - U

Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame

V - speed of object in the unprimed frame

U - speed of primed frame with respect to the unprimed frame

Here we have:

V = -15 m/s speed of ball in the moving frame (the truck)

U = -20 m/s speed of primed (rest) frame with respect to moving frame

So V' = -15 - (-20) = 5 m/s

It may help if you draw a vector representing the moving frame and then add

a vector representing the speed of the ball in the moving frame.

3 0

3 years ago

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6 = V/0.0001

V = 2.0 × 10^6 * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q = 1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0

2 years ago

A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p

Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d = 2.0 mm = 2 x 10⁻³ m

magnitude of the electric field = 200 kN/C

Capacitance of the capacitor is calculated as follows;

$C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F$

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

4 0

3 years ago