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Mashcka [7]
3 years ago
14

Describe the difference between particle motion in solids and liquids.

Physics
2 answers:
4vir4ik [10]3 years ago
5 0

Answer:

Particles in a: gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

i hope this helps your answer

jenyasd209 [6]3 years ago
3 0

Answer:

liquids vibrate, move about, and slide past each other. solids vibrate but generally do not move from place to place.

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The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

K_f =\frac{1}{2}mv^2

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J

5 0
3 years ago
Read 2 more answers
Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavel
Sergio [31]

Answer:

a) y = 2.4 x 10⁻³ m = 0.24 cm

b) y = 3.2 x 10⁻³ m = 0.32 cm

Explanation:

The formula of Young's Double Slit experiment will be used here:

y = \frac{\lambda L}{d}\\\\

where,

y = distance between dark spots = ?

λ = wavelength

L = distance of screen = 2 m

d = slit width = 4 x 10⁻⁴ m

a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}

<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>

<u></u>

a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}

<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>

7 0
2 years ago
Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2
liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

7 0
2 years ago
The harder I push an object the _______ it will travel.
lions [1.4K]

It's b, because the more force an object it is given the harder it will be for it to slow down.

8 0
3 years ago
Read 2 more answers
Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
d1i1m1o1n [39]
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
3 0
3 years ago
Read 2 more answers
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