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3 years ago

What state of matter has a defined volume but undefined shape?

Physics

2 answers:

slavikrds [6]3 years ago

6 0

LIQUID is that state............

grin007 [14]3 years ago

4 0

Liquid and Gases is the correct answer

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What is the specific heat of silver if the temp ig a 15.4g sample of silver is increased frin 20 degrees celsius to 31.2 degrees

Genrish500 [490]

heat = mass x spec heat x temp rise

40.5=15.4x10^-3xspec heatx11.2

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3 years ago

Difference between moment and displacement

harkovskaia [24]

Displacement is the difference between final position and initial position.

Momentum is the quantity of motion contained by an object.

It is the product of mass and velocity.

7 0

3 years ago

A particle travels in a circle of radius 76 cm and completes one revolution in 4.5 s. What is the centripetal acceleration of th

Dimas [21]

r = radius of the circle traveled by the particle = 76 cm = 0.76 m

T = time period of revolution for the particle = 4.5 s

w = angular velocity of the particle

angular velocity of the particle is given as

w = 2π/T

inserting the values

w = 2 (3.14)/4.5

w = 1.4 rad/s

a = centripetal acceleration of the particle in the circle

centripetal acceleration is given as

a = r w²

inserting the values

a = (0.76) (1.4)²

a = 1.5 m/s²

3 0

3 years ago

Read 2 more answers

A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and

MariettaO [177]

Answer:

$f=140\ N$

Explanation:

Given:

mass of the object on a horizontal surface, $m=50\ kg$

coefficient of static friction, $\mu_s=0.3$

coefficient of kinetic friction, $\mu_k=0.2$

horizontal force on the object, $F=140\ N$

Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:

$F_s=\mu_s.N$

where:

$N=$ normal force of reaction acting on the body= weight of the body

$F_s=0.3\times (50\times 9.8)$

$F_s=147\ N$

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

$f=140\ N$

8 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago