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4 years ago

Can someone please help with this page

Physics

1 answer:

Allisa [31]4 years ago

3 0

I have a strong hunch that if you read through pages 184 to 187 in the book,
you'll find each of these statements there, with no blank spaces.

1. negative
2. static
3. electrons
4. repel
5. attract
6. static
7. electric current
8. closed or complete
9. open

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PART ONE

Arte-miy333 [17]

Explanation:

Pressure = force / area

P = (68 kg × 9.8 m/s²) / (2 × (0.04 m)²)

P = 208,250 Pa

Converting to psi:

P = 208,250 N/m² × (0.225 lbf/N) × (0.0254 m/in)²

P = 30.2 psi

4 0

4 years ago

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of t

9966 [12]

Answer: a) 0.11 Amp b) 0.000429 Joules c) 0.000514 watt

Explanation:

a)current

I=V/r I=6.0/550 =1.1*10^-2 A

b) energy

E= 1.1*10^-2 x (6.0) x 0.00065s(<-- remember to convert ms to seconds) = 0.000429J

C) Average power

P = (0.000429 x 72) / 60

P = 0.000514 W

6 0

3 years ago

What is most often given a value of zero to describe an object position on a srtaight line

Vsevolod [243]

Reference point is most often given a value of zero to describe an object's position on a straight line

6 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

What is the static friction force and kinetic friction for the image

jasenka [17]

According to the plot, static friction force has a maximum magnitude of around 3.0 N, and kinetic friction has a magnitude of about 1.5 N.

The plot appears to be telling you the force required to get the yellow block moving along the table. If one applies less than 3.0 N of force, the block remains motionless. But as soon as it starts to slide, one need only apply 1.5 N of force to keep it moving (presumably at a constant speed).

8 0

3 years ago