What happens when an object with a lower density is placed in a container with an

If Mercury, moved two orbital paths closer to the sun: what would be different? List serval differences, ideas.......

Well first of all, when it comes to orbits of the planets around
the sun, there's no such thing as "orbital paths", in the sense
of definite ("quantized") distances that the planets can occupy
but not in between. That's the case with the electrons in an atom,
but a planet's orbit can be any old distance from the sun at all.

If Mercury, or any planet, were somehow moved to an orbit closer
to the sun, then ...

-- its speed in orbit would be greater,

-- the distance around its orbit would be shorter,

-- its orbital period ("year") would be shorter,

-- the temperature everywhere on its surface would be higher,

-- if it has an atmosphere now, then its atmosphere would become
less dense, and might soon disappear entirely,

-- the intensity of x-rays, charged particles, and other forms of
solar radiation arriving at its surface would be greater.

5 0

3 years ago

Radar uses radio waves of a wavelength of 2.2 m . The time interval for one radiation pulse is 100 times larger than the time of

liq [111]

Answer:

The shortest distance to an object that this radar can detect would be

111 m

Explanation:

The shortest distance is the minimum distance that would be detected by the radar, The time of oscillation can be obtained thus;

T = 1/f ..........1

but f= v/λ substituting f into equation 1 we have;

T = λ/v...............2

Where λ is the wavelength = 2.2 m

v is the velocity of light since the radar is an electromagnetic wave

= 3 x $10^{8}$ m/s

T = 2.2 m / 3 x $10^{8}$ m/s

T = 7.33 x $10^{-9}$ s

Calculating the time interval required by the pulse

Since the interval of the pulse (t) is 100 times greater than the period of oscillation, the time of the pulse is expressed as;

t = 100 x T

t = 100 x 7.33 x $10^{-9}$ s

t = 7.33 x $10^{-7}$ s

Therefor the time of the pulse is 7.33 x $10^{-7}$ s

Calculating for the shortest distance of the radar

The shortest distance of the radar can be obtained using the equation below;

λ_s = (t/2) x v ............3

Substituting into equation 3 we have

λ_s = (7.33 x $10^{-7}$ /2) x 3 x $10^{8}$ m/s

λ_s = 111 m

Therefore the shortest distance to an object that this radar can detect would be 111 m

8 0

4 years ago

Determine the average acceleration for x(t)=19t^2+7t^3 for a time interval between 3 and 9 seconds

AleksandrR [38]

Answer:

290

Explanation:

Average acceleration is the change in velocity over change in time.

First, find the velocity by taking the derivative of position.

v(t) = dx/dt

v(t) = 38t + 21t²

At t = 3 and t = 9:

v(3) = 303

v(9) = 2043

So the average acceleration is:

a = Δv / Δt

a = (2043 − 303) / (9 − 3)

a = 290

Use appropriate units.

7 0

3 years ago

Calculate the time of fight for a horizontally launched projectile from a height of 20m above the ground with an initial velocit

Hoochie [10]

Answer:

0.5sec

Explanation:

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Method one

Using the second law of motion

H=ut-1/2gt^2

20=5t-1/2×10×t^2

20=5t-5t^2

dh/dt = 5-10t

where any constant is zero therefore the 20 is zero

5-10t=0

Collect like terms

-10t= -5

t=1/2 = 0.5sec

2nd method

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Using the time taken formula

t=u/g

t=5/10

t=0.5sec

4 0

3 years ago

Balance is the body's ability to maintain ____________ and stability.

aev [14]

Answer:

Equilibrium or coordination

Explanation:

The body's equilibrium is it's way to maintain balance, and perfect distribution of weight. When no force is acting to make a body move in a line, the body is in translational equilibrium; when no force is acting to make the body turn, the body is in rotational equilibrium.

8 0

4 years ago