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Stells [14]
1 year ago
12

What mass of silver has a volume of 150 cm3?

Physics
1 answer:
Lena [83]1 year ago
3 0

ANSWER:

1573.5 g

STEP-BY-STEP EXPLANATION:

Given:

Volume (V) = 150 cm³

We have that the density of silver is 10.49 g/cm³, therefore, we calculate its mass, as follows:

\begin{gathered} d=\frac{m}{V} \\  \\ \text{ We replacing:} \\  \\ 10.49=\frac{m}{150} \\  \\ m=10.49\cdot150 \\  \\ m=1573.5\text{ g} \end{gathered}

The mass of silver is 1573.5 grams

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What happens to the rock structure during each of type of change?
EastWind [94]
Previous rocks melt and collide and to form igneous rocks.
Igneous rocks disintegrate due to weather disruptions and get carried away by water, where they form sedimentary rock strata by lithification.
Igneous and sedimentary change by heat and pressure to form metamorphic rocks.
Metamorphic rocks melt and become igneous rocks.
3 0
3 years ago
A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p
saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286

0.12\omega =\cos^{-1}0.4286

0.12\omega = 1.13

\omega = 9.4

The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

5 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
An eccentric inventor attempts to levitate by first placing a large negative charge on himself and then putting a large positive
GuDViN [60]

Answer:

shirt is little attached to the body, it can come off and fly away

Explanation:

In electrostatics, charges of different signs attract and charges of the same sign repel.

In this case, when a negative charge is placed on it, both the inventor and the shirt are charged, therefore there is a repulsive force, also there is an attraction between the positive charge of the roof attracts the negative charge, such as the shirt. of weak the two forces not greater than the resistance of the walk.

As the shirt is little attached to the body, it can come off and fly away

3 0
2 years ago
1.) A stone falls from rest from the top of a cliff.
KengaRu [80]

Answer:

Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716

     0 = 21.56t - 71.148

t = 71.148/21.56 = 3.3 s

h = ½(9.8)3.3² = 53.361 = 53 m

or

h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

4 0
2 years ago
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