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kompoz [17]
3 years ago
10

A ball whose mass is 1.9 kg is suspended from a spring whose stiffness is 8.0 N/m. The ball oscillates up and down with an ampli

tude of 17 cm. Part 1 (a) What is the angular frequency? ω= radians/s the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.
Physics
1 answer:
MArishka [77]3 years ago
4 0

Answer:

2.05 radians/s

Explanation:

This is a simple harmonic motion. The angular frequency of a loaded spring is given by

\omega = \sqrt{\dfrac{k}{m}}

where k is the spring constant and m is the mass on the spring.

Using the known values,

\omega = \sqrt{\dfrac{8}{1.9}} = 2.05

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Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h
vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}

\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

8 0
3 years ago
A metal ball with a charge of -8 C is brought in contact with a similar metal ball that had a charge of +2 C. How much charge wi
laila [671]
I think the answer is -6
5 0
2 years ago
The mass of a rocket decreases as it burns through its fuel. If the rocket engine produces constant force (thrust), how does the
Serjik [45]

Answer:

it increases-

Explanation:

When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion

F= ma  here F is constant this means that   ma= constant

⇒ m= F /a    this implies that mass is inversely proportional to  acceleration.

its means when the mass decreases the acceleration must increase. hence the acceleration increases

6 0
3 years ago
If anyone knows please anwser
Nonamiya [84]

Answer:

C.

Explanation:

4 0
3 years ago
Read 2 more answers
2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas
Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

ds=\dfrac{dQ}{dt}

At constant pressure,

dQ=C_{p}dt

\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}

\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}

Put the value into the formula

14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})

\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}

0.693=ln\dfrac{T_{2}}{T_{1}}

ln2=ln\dfrac{T_{2}}{T_{1}}

T_{2}=2T_{1}...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

\Delta Q=C_{p}\Delta T

Put the value into the formula

6236=2.5\times8.3144(T_{2}-T_{1})

6236=20.786(T_{2}-T_{1})

T_{2}-T_{1}=\dfrac{6236}{20.786}

T_{2}-T_{1}=300

Put the value of T₂

2T_{1}-T_{1}=300

T_{1}=300\ K

Put the value of T₁ in equation (I)

T_{2}=2\times300

T_{2}=600\ K

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0
3 years ago
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