Relative to the distance of an object in front of a plane mirror, how far behind the mirror is the image?

A machine shop worker reports the mass of an aluminum cube as 176 g. If one side of the cube measures 4 cm, what is the density

Zarrin [17]

-- Since it's a cube, its length, width, and height are all the same 4 cm .

-- Its volume is (length x width x height) = 64 cm³ .

-- Density = (mass) / (volume)

= (176 g) / (64 cm³)

= 2.75 gm/cm³ .

6 0

2 years ago

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goblinko [34]

Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

6 0

2 years ago

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

a 4kg metal block absorbs 5000j of energy and increases to a temperature of 22°c. the metal has a specific heat capacity of 250j

e-lub [12.9K]

Answer:

17 °C

Explanation:

From specific Heat capacity.

Q = cm(t₂-t₁)................. Equation 1

Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.

make t₁ the subject of the equation

t₁ = t₂-(Q/cm)............... Equation 2

Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c

Substitute into equation 2

t₁ = 22-[5000/(4×250)

t₁ = 22-(5000/1000)

t₁ = 22-5

t₁ = 17 °C

6 0

2 years ago

A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the

Serjik [45]

The elastic potential energy of a spring is given by
$U= \frac{1}{2}kx^2$
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
$W=-\Delta U = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$

In our problem, initially the spring is uncompressed, so $x_i=0$ . Therefore, the work done by the spring when it is compressed until $x_f$ is
$W=- \frac{1}{2}kx_f^2$
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.

8 0

3 years ago