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3 years ago

What is the difference between sodium atom and a sodium ion

1 answer:

7 0

Pretty much any element(in your case sodium) contain these properties.

Atoms can be an ion, but not all ions are atoms. The difference between an atom and an ion has to do with net electrical charge. An ion is a particle or collection of particles with a net positive or negative charge. ... A stable atom contains the same number of electrons as protons and no net charge

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A 1 pound bucket rests on a table. What is the support force exerted on the bucket by the table? What is the support force when

katrin2010 [14]

This question involves the concepts of equilibrium and Newton's third law of motion.

The support force will be "1 pound" for the empty bucket and the support force will be "6 pounds" after pouring water into it.

According to the condition of equilibrium, the sum of forces acting on a stationary object must be zero. Hence, the support force of the table will be equal to the total mass of the bucket.
According to Newton's Third Law of Motion every action force has an equal but opposite reaction force. Hence, the support force will be a reaction force to the weight of the bucket.

Therefore, the support force in each case will be equal to the total mass of the bucket:

Case 1 (empty bucket):

support force = 1 pound

Case 1 (water poured):

support force = 1 pound + 5 pound

support force = 6 pound

Learn more about equilibrium here:

brainly.com/question/9076091

8 0

3 years ago

When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g

mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=m $c_p$ dT

Where m is the mass $c_p$ = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

$T_1$ =25°C

$T_2$ we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25- $T_2$ )

25- $T_2$ =22

$T_2$ =3°C

7 0

3 years ago

Read 2 more answers

A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and

s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r = m x g x μ

v^2 / r = g x μ .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r = g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0

3 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 146 cm and makes an angl

svetlana [45]

Answer:

B = 191.26 cm

θ = -14.73°

Explanation:

given,

magnitude of the first displacement(A) = 146 cm

at an angle of 124°

resultant magnitude = 137 cm

and angle made with x-axis by the resultant(R) = 32.0°

component of A in X and Y direction

A x = A cos θ = 146 cos 120° = -73 cm

A y = A sin θ = 146 sin 120° = 126.4 cm

now component of resultant in x and y direction

R x = 137 cos 35°

= 112.2 cm

R y = 137 sin 35°

= 78.6 cm

resultant is the sum of two vectors

R = A + B

R x = A x + B x

B x = 112.2 - (-73) = 185.2 cm

B y = R y - A y

B y = 78.6 - 126.4 = -47.8 cm

magnitude of B

B = $\sqrt{B_x^2+B_y^2}$

B = $\sqrt{185.2^2+-47.8^2}$

B = 191.26 cm

angle $\theta = tan^{-1}\dfrac{-47.8}{185.2}$

θ = -14.73°

6 0

4 years ago

You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho

timurjin [86]

Answer:

$\omega'=19.419\ rev.s^{-1}$

Explanation:

Given:

angular speed of rotation of friction-less platform, $\omega=5.1\ rev.s^{-1}$

moment of inertia with extended weight, $I=9.9\ kg.m^2$

moment of inertia with contracted weight, $I'=2.6\ kg.m^2$

Now we use the law of conservation of angular momentum:

$I.\omega=I'.\omega'$

$9.9\times 5.1=2.6\times \omega'$

$\omega'=19.419\ rev.s^{-1}$

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

5 0

3 years ago