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Wittaler [7]
3 years ago
12

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

then pulled down 7.5 cm and released. The ball makes 31 oscillations in 15 s seconds.1) What is its the mass of the ball?Express your answer using two significant figures.m = _____ g2) What is its maximum speed?Express your answer using two significant figures.= _____ cm/s
Physics
1 answer:
olga nikolaevna [1]3 years ago
5 0

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Total number of oscillation per unit time is called frequency of oscillation. Here, f=\dfrac{31}{15}=2.06\ Hz

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{15}{4\pi^2\times 2.06^2}

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

v_{max}=A\times \omega

v_{max}=A\times 2\pi f

v_{max}=0.075\times 2\pi \times 2.06

v_{max}=0.970\ m/s

v_{max}=97\ cm/s

Hence, this is the required solution.

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mestny [16]

Answer:

The correct answer is - Plantae.

Explanation:

Drosera m<em>agnifica</em> is discovered in 2015 for the first time and the characteristics this organism's cell shows are -

- permanent vacuoles

- surrounded by cellulose layer

Vacuoles are present in both Plantae and Animalia kingdom of the eukaryotic organism but in animal cells, there are small and numerous vacuoles present and they are not permanent whereas in plant cells vacuoles are present permanently.

The cell of an animal cell has no surrounding layer other than cell membrane while in the plant cell there is a supporting and protecting layer of cellulose cell wall present.

On the basis of the given characteristics, it is confirmed that the Drosera magnifica belongs to Plantae kingdom.

3 0
3 years ago
all of the following answer choices are ideas from daltons atomic theory. which one of them do we now know is not true?
dem82 [27]
Did you forget to include the options?
5 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

4 0
3 years ago
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Lelechka [254]
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6 0
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Sladkaya [172]

Answer:

Explanation:

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