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3 years ago

Physics is a branch of science which deals with........... in relation to.................

2 answers:

5 0

Physics is a branch of science which deals with structure of matter and how the fundamental constituents of the universe interact. 

hope it helps

dybincka [34]3 years ago

3 0

Answer:

Physics is the branch of science that deals with the structure of matter and how the fundamental constituents of the universe interact.

i think so

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The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i

fomenos

To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

We can express this also as,

$log_{10}(10^6)$

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

$6*log_{10}(10)$

Since the definition of the base logarithm 10 of 10 is equal to 1 then

$6*1=6$

The value of the given logarithm is equal to 6

8 0

3 years ago

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

3 years ago

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

Δx = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

6 0

3 years ago

If the distance between two objects decreased, what would happen to the force of gravity between them?

evablogger [386]

Answer: A- It would increase

Explanation:

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the attraction force exerted between both objects

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both objects

$r$ is the distance between both objects

As we can see, the gravity force is directly proportional to the mass of the bodies or objects and inversely proportional to the square of the distance that separates them.

In other words:

<h2>If we decrease the distance between both objects, the gravitational force between them will increase. </h2>

6 0

3 years ago

Read 2 more answers

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$