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3 years ago

A train moves from rest to a speed of 25 m/s in 30.0 seconds . What is the acceleration

1 answer:

7 0

When a car moves from rest then the initial velocity,
$u = 0m {s}^{ - 1}$

The final velocity,

$v = 25m {s}^{ - 1}$
and time,
$t = 30s$
Using the equation,
$v = u + at$
We have,

$25 = 0 + a(30)$
$25 = 30a$
$\frac{25}{30} = a$
$\therefore \: a = \frac{5}{6} m {s}^{ - 2}$

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A square loop of side length a =4.5 cm is placed a distance b = 1.1 cm from a long wire carrying a current that varies with time

sergey [27]

Answer:

a)Current will flow perpendicularly.

b)Magnitude of flux will be 2.987 N m2 C−1

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3 years ago

Read 2 more answers

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

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3 years ago

Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the

grandymaker [24]

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

3 0

3 years ago

Read 2 more answers

A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North

ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

$magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km$

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km

3 0

3 years ago

Calcium oxide write this in a formula

kykrilka [37]

The formula for calcium oxide is CaO.

4 0

3 years ago