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3 years ago

What will happen two temperatures if you increase particle motion

1 answer:

8 0

If the temperature is increased the particles gain more kinetic energy or vibrate faster. This means that they move faster and take more space.

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gregori [183]

Answer:

1. Revolve around a point

2. Formed from dust and gas particles

3. Exoplanets and associated star orbit a common center of mass

4. Composed of gases found in Jupiter and Saturn

3 0

3 years ago

A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi

S_A_V [24]

Answer:

10.2 Watt

Explanation:

$N$ = number of turns in flat coil = 160

$A$ = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

$B$ = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

$t$ = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

$E = \frac{- N A \Delta B}{t}$

$E = \frac{- (160) (0.20) (- 0.80)}{2}$

$E$ = 12.8 volts

$R$ = Resistance of the coil = 16 Ω

Power is given as

$P = \frac{E^{2}}{R}$

$P = \frac{(12.8)^{2}}{16}$

$P$ = 10.2 Watt

6 0

3 years ago

A sprinter who is running a 200-m race travels the second 100 m in much less time than the first 100m because

Harrizon [31]

Answer:

He is warmed up now

Explanation:

His muscles are better and stretched now

3 0

3 years ago

Read 2 more answers

If the distance between the first order maximum and the tenth order maximum of a double-slit pattern is 18 mm and the slits are

Setler79 [48]

Answer:

Wavelength of light is 600 nm

Explanation:

Given

Distance between the first order maximum and the tenth order maximum of a double-slit pattern = 18 mm

Separation between the slits = 0.15 mm

Distance of screen from the slits = 50 cm

Wavelength

$= \frac{18*10^{-3} * 0.15 *10^{-3}}{0.50*9} \\= 6 *10^{-7}\\= 600$ nm

4 0

3 years ago

How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume

Svetach [21]

Answer:

$9.56\cdot 10^{-7} C$

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma=\frac{Q}{A}$ is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

$A=L^2=(0.19)^2=0.036m^2$

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

$E=3\cdot 10^6 N/C$

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

$E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C$

7 0

4 years ago