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nikitadnepr [17]
3 years ago
7

A student uses a spring to launch a marble vertically in the air. The mass of the marble is

Physics
1 answer:
Fed [463]3 years ago
5 0

Answer:

the maximum height reached by the marble is 11.84 m.

Explanation:

Given;

mass of the marble, m = 0.02 kg

extension of the string, x = 0.08 m

force applied to the string, F = 58 N

Apply the principle of conservation of energy;

elastic potential energy = gravitational potential energy

¹/₂fx = mgh

h = \frac{Fx}{2mg} \\\\h = \frac{58 \ \times \ 0.08}{2 \ \times \ 0.02 \ \times \ 9.8} \\\\h = 11.84 \ m

Therefore, the maximum height reached by the marble is 11.84 m.

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What is the increase in pressure required to decrease volume of mercury by 0.001%
REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0
3 years ago
Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 2.4
NeX [460]

When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength

now we will have

\frac{\lambda}{2} = \frac{yd}{L}

now we know that

y = 0.20 m

d = 2.4 m

L = 10 m

now here we have

\frac{\lambda}{2} = \frac{0.20\times 2.4}{10}

\lambda = 0.096 m

now frequency of wave is given as

f = \frac{v}{\lambda}

f = \frac{343}{0.096} = 3573 Hz

5 0
3 years ago
A pyramid was built with approximately 2.3 million stone blocks, each weighing 2.4 tons (1 ton = 2,000 lbm). Find the mass of th
pogonyaev

Total weight in tons

\\ \sf\longmapsto 2.3\times 10^6\times 2.4

\\ \sf\longmapsto 5.52\times 10^6

\\ \sf\longmapsto 5520000ton

  • 1ton=1000kg

\\ \sf\longmapsto 5520000\times 1000

\\ \sf\longmapsto 5520000000kg

\\ \sf\longmapsto 5.52\times 10^9Kg

4 0
3 years ago
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by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th
xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}

\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}

\displaystyle F'=9k\frac{q_1q_2}{d^2}

\displaystyle F'=9F

The force is now 9 times the original force

8 0
3 years ago
Are dimensionless quantities always unitless​
timofeeve [1]

<em>Important thing is that all unitless quantity is dimensionless quantity. .</em><em>A</em><em> dimensionless physical quantity may have an unit</em>

4 0
3 years ago
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