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KATRIN_1 [288]
4 years ago
12

What happens to a substance when it dissolves

Physics
1 answer:
11Alexandr11 [23.1K]4 years ago
4 0

Answer:

When a substance dissolves, the solute breaks up from larger crystal molecules into much smaller groups or individual molecules.

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What is the energy of a photon of ultraviolet radiation with a frequency of 4.4 × 1015 Hz? Planck’s constant is 6.63 × 10–34
Hunter-Best [27]

Answer:

2.9\cdot 10^{-18} J

Explanation:

The energy of a photon is given by:

E=hf

where

h is the Planck constant

f is the frequency of the photon

In this problem, we have:

h = 6.63\cdot 10^{-34} Js

f=4.4\cdot 10^{15} Hz is the frequency of the photon

Substituting into the equation, we find:

E=(6.63\cdot 10^{-34} Js)(4.4\cdot 10^{15} Hz)=2.9\cdot 10^{-18} J

4 0
3 years ago
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Can someone help me with this
Monica [59]

Answer:

yes

Explanation:

what do u need help with

6 0
3 years ago
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance x from the 50
Rom4ik [11]

Answer:

x=0.01457 m

Explanation:

From parallel axis theorem

I=Icm+mh²

where h=x

The rotational inertia about its center of mass is

Icm=mL²/12

where L=1.0 m

Thus T=4.8s  we obtain

T=2\pi \sqrt{\frac{mL^{2}/12+mx^{2}  }{mgx} }\\ T=2\pi \sqrt{\frac{L^{2} }{12gx}+x/g }\\ T^{2}=4\pi^{2}(\frac{L^{2} }{12gx}+x/g )/x\\ T^{2}x=\frac{\pi^{2} L^{2} }{3g}+(\frac{4\pi^{2} }{g})x^{2}\\   0=(\frac{4\pi^{2}  }{g} )x^{2}-(T^{2} )x+(\frac{\pi^{2}L^{2}   }{3g} )\\ 4.03x^{2}-23.04x+0.335=0

After Solving this quadratic we get

x₁=5.702 m

x₂=0.01457 m

One of the solution is an impossible value for x (x=5.70m is greater than L)

So we choose the other one

x=0.01457 m

3 0
4 years ago
A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh
Kazeer [188]

Answer:

a)t = H/v_0

b)H = v_0^2/g

Explanation:

Let the first ball throw be the point of reference, we can have following the equation of motion:

1st ball: h_1 = v_0t - gt^2/2

2nd ball: h_2 = H - gt^2/2

a)When the 2 balls collide they are at the same spot at the same time:

h_1 = h_2

v_0t - gt^2/2 = H - gt^2/2

v_0t = H

t = H/v_0

b) The first ball is at its highest point when v = 0. That is

t = v_0/g

After this time, the 2 balls would have traveled through a distance of

h_1 = v_0t - gt^2/2 = v_0^2/g - v_0^2/2g

h_2 = gt^2/2 = v_0^2/2g

SinceH = h_1 + h_2 we can solve for H

H = v_0^2/g - v_0^2/2g + v_0^2/2g = v_0^2/g

3 0
3 years ago
A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find th
Andre45 [30]

Answer:

0.026 V

Explanation:

Given that,

Inductance of the coil, L = 6 mH

The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s

We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :

\epsilon=L\dfrac{dI}{dt}\\\\=6\times 10^{-3}\times \dfrac{1.5-0.2}{0.3}\\\\=0.026\ V

So, the magnitude of induced emf is 0.026 volts.

5 0
3 years ago
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