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zavuch27 [327]
2 years ago
5

Which state of matter has a define shape and definite volume

Physics
2 answers:
Katena32 [7]2 years ago
8 0
<h3>Answer:</h3><h2>Solids Only</h2><h3>Explanation:</h3>

Off all the states of matter, only solids have a defined shape, due to stronger bonds between its molecules/atoms. Unlike solids, Liquids and gasses take the shape of whatever container they are placed into. This, again, is due to weaker linkages between the atoms/molecules of gasses and liquids.

geniusboy [140]2 years ago
3 0

A solid has a definite shape and volume


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Two 8.0 Ω lightbulbs are connected in a 12 V parallel circuit. What is the power of both glowing bulbs?
kati45 [8]

Answer:

96w

Explanation:

p=Iv..where v=12 and I=8.0

8 0
2 years ago
A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67
aliya0001 [1]
Acceleration =velocity /time
=42/7
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5 0
3 years ago
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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
Nezavi [6.7K]

Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0  x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12    (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))  

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR)  --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides  

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x 10^{-9  

t = 23.9ns

Thus, it took 23.9ns  for the potential difference between the deflection plates to reach 75 volts

6 0
3 years ago
Read 2 more answers
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
A 0.07-kg lead bullet traveling 258 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted t
Lubov Fominskaja [6]

Answer:

temperature change is 262.06°K

Explanation:

given data

mass = 0.07 kg

velocity = 258 m/s

to find out

what is its temperature change

solution

we know here

heat change Q is is equal to kinetic energy that is

KE = 0.5 × m× v²   ...........1

here m is mass and v is velocity

KE = 0.5 × 0.07 × 258²

KE = 2329.74 J

and we know

Q = mC∆t     .................2

here m is mass and ∆t is change in temperature and C is 127J/kg-K

so put here all value

2329.74 = 0.07 × 127 × ∆t

∆t = 262.06

so temperature change is 262.06°K

7 0
3 years ago
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