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3 years ago

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The initial velocity of a micro van is 15 m/s. It gains a velocity of 40 ms in 10 seconds. Calculate the average velocity and ac

celeration of the [Ans: 27.5 m, 25 m/s d/ A racing car starts from rest. If it gains an acceleration of 5m/s in 10 seconds, calculate the final velocity. (Ans: 50 m van.

1 answer:

PSYCHO15rus [73]3 years ago

5 0

Answer:

${ \bf{average \: velocity = \frac{15 + 40}{2}}} \\ = 27.5 \: {ms}^{ - 1} \\ { \bf{acceleration = \frac{v - u}{t} }} \\ = \frac{40 - 15}{10} \\ = 2.5 \: {ms}^{ - 2} \\ \\ { \tt{second \: qn : }} \\ { \bf{final \: velocity =u + at }} \\ v = 0 + (5 \times 10) \\ = 50 \: {ms}^{ - 1}$

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

Diamond has a high index of refraction at about 2.4, which helps account for its sparkle. How fast does light travel through a d

wariber [46]

Answer:

$1.25 \cdot 10^8 m/s$

Explanation:

The speed of light through a medium is given by:

$v=\frac{c}{n}$

where

$c = 3 \cdot 10^8 m/s$ is the speed of light through vacuum

n is the index of refraction of the material

In this case, diamond has a refractive index of n = 2.4, so the speed of light in diamond is

$v=\frac{3\cdot 10^8 m/s}{2.4}=1.25\cdot 10^8 m/s$

6 0

3 years ago

Write the following in expanded form. 4.35 x 10^6

EleoNora [17]

4.35x10x10x10x10x10x10 if you wanted complete expanded form it would be
4+.3+.05x10x10x10x10x10x10

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4 years ago

Two strong permanent magnets are placed one inch apart. Which two fundamental forces will most affect them?

Anuta_ua [19.1K]

The answer i got was c. gravity and strong
i hope it helps

3 0

3 years ago

Molybdenum has the BCC crystal structure, has a density of 10.22 g cm-3 and an atomic mass of 95.94 g mol-1. What is the atomic

puteri [66]

Answer:

atomic concentration = 2 atoms/unit cell

lattice parameter: a= 3.22 x 10⁻¹⁰ m

atomic radius: r= 1.39 x 10⁻¹⁰m

Explanation:

The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell

In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:

Vc= a³= $\frac{(95.94 g/mol) x (2 atoms/unit cell)}{(10.2 g/cm^{3}) x (6.023 x 10^{23} atoms/mol) }$

a³= 3.12 x 10⁻²³ m³

⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m

Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:

a= $\frac{4r}{\sqrt{3}}$

⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m

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3 years ago