Please answer my question

A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, which has an altitude of 10 m. There is no fricti

damaskus [11]

Answer:

the speed of the block when it reaches point B is 14 m/s

Explanation:

Given that:

mass of the block slides = 1.5 - kg

height = 10 m

Force constant = 200 N/m

distance of rough surface patch = 20 m

coefficient of kinetic friction = 0.15

In order to determine the speed of the block when it reaches point B.

We consider the equation for the energy conservation in the system which can be represented by:

$\dfrac{1}{2}mv^2=mgh$

$\dfrac{1}{2}v^2=gh$

$v^2=2 \times g \times h$

$v^2=2 \times 9.8 \times 10$

$v=\sqrt{2 \times 9.8 \times 10$

$v=\sqrt{196$

v = 14 m/s

Thus; the speed of the block when it reaches point B is 14 m/s

7 0

3 years ago

Describe the main distinguishing features of spiral, elliptical, and irregular galaxies.

hram777 [196]

Answer:

Spiral galaxies consist of a flat, rotating disk of stars, gas and dust, and a central concentration of stars known as the bulge. These are surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Elliptical galaxies have smooth, featureless light-profiles and range in shape from nearly spherical to highly flattened, and in size from hundreds of millions to over one trillion stars. In the outer regions, many stars are grouped into globular clusters. Most elliptical galaxies are composed of older, low-mass stars, with a sparse interstellar medium and minimal star formation activity They are often chaotic in appearance, with neither a nuclear bulge nor any trace of spiral arm structure. Collectively they are thought to make up about a quarter of all galaxies.

irregular galaxies were once spiral or elliptical galaxies but were deformed by gravitational action. they are shapeless.

5 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago

The weather that characterizes an area is the O latitude of that area O barometric pressure of that area climate of that area O

Nikolay [14]

Answer:

climate of that area

Explanation:

3 0

2 years ago

A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string i

Firdavs [7]

Answer:

a

The tension in the string is $T = 0.85 N$

b

The new balance reading is $M_b = 885.86 g$

Explanation:

From the question we are told that

The mass of the beaker of water is $m = 875 .0g$

The diameter of the copper ball is $d = 2.75 cm = \frac{2.75}{100} = 0.0275 m$

There are two forces acting on the copper ball

The first is the Buoyant force of the water pushing it up which is mathematically represented as

$F = \rho V g$

Where $\rho$ is the density of water which has value of $\rho = 1000 kg/m^3$

g is the acceleration due to gravity $g= 9.8 \ m/s^2$

$V$ is the volume of water displaced by the copper ball which is mathematically evaluated as

$V = \frac{4}{3} \pi r^3$

The radius r is $r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m$

Substituting value

$V = \frac{4}{3} * 3.142 * (0.01375)^3$

$V = 1.08 9 *10^{-5 } m^3$

Substituting for F

$F = 1000 * 1.089 *10^{-5} * 9.8$

$F = 0.1067 N$

The second force is the weight of the copper ball which is mathematically represented as

$W_c = mg$

Now m is the mass which can be mathematically evaluated as

$m = \rho_c * V$

Where is the density of copper with value of $\rho_c = 8960 kg /m^3$

So $m = 8960 * 1.089 *10^{-5}$

$m = 0.0976$

So the weight of copper is

$W_c = 0.09756 * 9.8$

$W_c = 0.956 N$

Now the tension the string would be mathematically evaluated as

$T = W_c - F$

So $T = 0.956 - 0.1067$

$T = 0.85 N$

From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance

Now the mass equivalent of this weight is mathematically evaluated as

$m_z = \frac{1.0645 }{9.8 }$

$m_z = 0.01086 kg$

Converting to grams

$m_z = 10.86 g$

So the new balance reading is

$M_b = 875.0 +10.86$

$M_b = 885.86 g$

5 0

3 years ago