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4 years ago

PLEASE HELP!!

Physics

1 answer:

kolezko [41]4 years ago

3 0

ATOMIC MASS- 3

PROTONS- 1

NEUTRONS- 4

ISOTOPE- 5

ATOMIC NUMBER- 2

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A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

goblinko [34]

Oh my lord lol I was do ready to help then I saw numbers

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3 years ago

Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.

Paraphin [41]

Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

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3 years ago

An observer sitting at a bus stop sees the bus drive by at 30 m/s to the east. He also sees a

Dafna1 [17]

Answer:

32 m/s

Explanation:

The speed of a bus is 30 m/s due East wrt the passenger

He also sees a passenger on the bus walking to the back at 2 m/s.

We need to find the passenger's velocity relative to the bus. As the observer sees that the bus and the passenger are moving in opposite direction. Let v is the relative velocity. So,

v = 30 m/s + 2 m/s

v = 32 m/s

Hence, the passenger's velocity relative to the bus is 32 m/s.

8 0

4 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and

WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g $/m^{2}$ = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = $\frac{1}{2L}$ $\sqrt{\frac{T}{M} }$

⇒ T = 4 $L^{2}$ $f^{2}$ M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × $2^{2}$ × $65^{2}$ × 0.01

= 4 × 4 ×4225 × 0.01

= 676 N

Tension of the wire is 676 N.

4 0

3 years ago