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2 years ago

10

Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What will the sp

ring scale read if the elevator is moving upward at constant speed?

1 answer:

zloy xaker [14]2 years ago

4 0

Explanation:

if the elevator is moving upward with the constant speed the spring scale will read 18 N which is the mass of each of the two blocks attached by separate springs to the scale at opposite ends.

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Daria was swimming in a friend’s pool yesterday, when she saw that a fly had landed in the water about 5 feet away from her. She

jasenka [17]

Answer:

Daria probably suffers from Entomophobia.

5 0

2 years ago

23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n

PSYCHO15rus [73]

Answer:

a) (95.4 i^ + 282.6 j^) N , b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

sin θ = Fjy / Fj

cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

T = 90 -32 = 58º

Fjy = Fj sin 58

Fjx = FJ cos 58

Fjx = 180 cos 58 = 95.4 N

Fjy = 180 sin 58 = 152.6 N

Andrea's force is

Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

Fx = Fjx

Fx = 95.4 N

Axis y

Fy = Fjy + Fa

Fy = 152.6 + 130

Fy = 282.6 N

F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

F² = Fx² + Fy²

F = √ (95.4² + 282.6²)

F = √ (88963)

F = 298.27 N

tan θ = Fy / Fx

θ = tan-1 (282.6 / 95.4)

θ = tan-1 (2,962)

θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

F' = 298.27 N

θ' = 180 + 71.3

θ = 251.4º

4 0

2 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

2 years ago

At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

vlabodo [156]

Answer:

$20.3125$ kJ/mol

Explanation:

$P_{i}$ = initial vapor pressure = 45.77 mm Hg

$P_{f}$ = final vapor pressure = 193.1 mm Hg

$T_{i}$ = initial temperature = 213.1 K

$T_{f}$ = final temperature = 243.7 K

$H$ = Heat of vaporization

Using the equation

$ln\left ( \frac{P_{f}}{P_{i}} \right ) = \left ( \frac{-H}{R} \right )\left ( \frac{1}{T_{f}} - \frac{1}{T_{i}}\right)$

$ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)$

$H = 20312.5$ J/mol

$H = 20.3125$ kJ/mol

8 0

2 years ago

An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni

Marysya12 [62]

When air is blown into the open pipe,

L = $\frac{nλ}{2}$

where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation

⇒λ= $\frac{2L} {n}$

Note here that n=1 is for fundamental, n=2 is first harmonic and so on..

⇒ third harmonic will be n=4

Given L=6m, n=4, solving for λ we get:

λ= $\frac{(2)*(6)}{4}$ =3m

Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:

c=f.λ Or f= $\frac{c}{λ}$

⇒f= $\frac{344}{3}$

≈115 Hz

8 0

2 years ago