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Brut [27]
4 years ago
8

Under ordinary conditions of temperature and pressure, the particles in a gas are Select one: a. closely packed. b. very far fro

m each other. c. held in fixed positions. d. able to slide past each other.
Physics
1 answer:
klemol [59]4 years ago
5 0

Answer:

The particle of a gas are very far apart from each other.

Explanation:

The states of matter comprises of solid, liquid and gas. The molecules of gas are free because they are not tightly held together by molecular force. The force between the molecules has been broken, hence the particles possesses the freedom to move about thereby possessing high kinetic energy (energy possessed by a body due to its motion). Since this molecules can easily move freely, they are always far apart from each other under ordinary temperature and pressure.

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A 50 kg wooden crate is slid across a flat surface for 60 meters by a force of 90.0 N. If the coefficient of sliding friction is
Alika [10]

The final speed of the crate is 6.3 m/s

Explanation:

Please note that there is a typo in the text of the question: the coefficient of sliding friction is 0.15, not 9.15.

First of all, we need to find the force of friction acting on the crate. This is given by:

F_f = \mu mg

where:

\mu=0.15 is the coefficient of friction

m = 50 kg is the mass of the crate

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

F_f = (0.15)(50)(9.8)=73.5 N

Now we can find the acceleration of the crate by using Newton's second law:

F-F_f = ma

where

F = 90.0 N is the force applied forward on the crate

F_f = 73.5 N is the force of friction, acting backward

a is the acceleration of the crate

Solving for a,

a=\frac{F-F_f}{m}=\frac{90.0-73.5}{50}=0.33 m/s^2 in the forward direction.

Finally, we can find the final speed of the crate by using suvat equations:

v^2-u^2 = 2as

where:

v is the final speed

u = 0 is the initial speed

a=0.33 m/s^2 is the acceleration

s = 60 m is the distance covered

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(0.33)(60)}=6.3 m/s

Learn more about friction, forces and Newton's second law here:

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4 years ago
A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitud
DaniilM [7]

Answer:

(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.

5 0
3 years ago
Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source
Alex_Xolod [135]

Answer:

The magnetic field strength  required to hold anti-protons, moving at 5.70 ✕ 10⁷ m/s in a circular path of 3.20 m in radius is 0.186 T.

Explanation:

To solve the question we note that the magnetic force on a moving charge is given by

F = q·v·B

Where

q = Charge

v = Velocity of the charge =5.70 ✕ 10⁷ m/s

B = Magnetic field

Based on Newton's second law,

Force = Mass, m × Acceleration, a = m × a

Where:

a = Acceleration

m = Mass of anti-proton = Mass of proton = 1.6726219 × 10⁻²⁷ kg

We note that for circular motion, acceleration a is given by

\alpha = \frac{v^{2} }{r} .

Where:

r = Radius = 3.20 m

Therefore, for the circular motion, force, F = \frac{m\cdot v^{2} }{r}

Equating the magnetic force equation to the circular force equation, we have

\frac{m\cdot v^{2} }{r} = q·v·B So that, we find B by making the subject of the formula as follows

B= \frac{m\times v^{2} }{r\times q \times v} . Which gives

B= \frac{m\times v }{r\times q} =  \frac{(1.6726219 \times 10^{-27}) \times (5.7\times 10^{7}) }{(3.20)\times (1.602\times 10^{-19}) } =  0.186 T

The magnetic field strength is

B = 0.186 T

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3 years ago
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Semi-truck because it has more inertia

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3 years ago
Please help me ?????
castortr0y [4]
3 because opposites attract 
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4 years ago
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