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3 years ago

How is the density of ice compare with the density of water? explain to me pplz

Physics

1 answer:

Viefleur [7K]3 years ago

3 0

Lower. Water expands on lower temperatures, meaning less molecules in 1 m3, thus making it less dense

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A radiographer stands six feet from the x-ray source when performing a portable chest exam and receives an exposure of 2 mGy. If

telo118 [61]

Answer:

I₂ = 8 mG

Explanation:

The intensity of a beam is

I = P / A

Where P is the emitted power which is 3) 3

Let's use index 1 for the initial position of r₁ = 6 ft and 2 for the second position r₂ = 3 ft

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of the beam if we assume that it is distributed either in the form of a sphere is

A₁ = 4π r²

We substitute

I₂ = I₁ (r₁ / r₂)²

I₂ = 2 (6/3)²

I₂ = 2 4

I₂ = 8 mG

3 0

3 years ago

Read 2 more answers

Three types of inattentional blindness include all of the following except __________.

mariarad [96]

The answer is: A. Choice Deafness

8 0

3 years ago

Read 2 more answers

How much work did the movers do (horizontally) pushing a 43.0-kg crate 10.4 m across a rough floor without acceleration, if the

Ilia_Sergeevich [38]

The crate is in equilibrium. Newton's second law gives

∑ F (vertical) = n - mg = 0

∑ F (horizontal) = p - f = 0

where

• n = magnitude of the normal force

• mg = weight of the crate

• p = mag. of push exerted by movers

• f = mag. of kinetic friciton, with f = 0.60n

It follows that

p = f = 0.60mg = 0.60 (43.0 kg) g = 252.84 N

so that the movers perform

W = p (10.4 m) ≈ 2600 J

of work on the crate. (The total work done on the crate, on the other hand, is zero because the net force on the crate is zero.)

8 0

3 years ago

Jason launches a model rocket with a mass of 2.0 kg from his spring-powered rocket launcher with a spring constant of 800 N/m. H

Arada [10]

Answer:

121 Joules

6.16717 m

Explanation:

m = Mass of the rocket = 2 kg

k = Spring constant = 800 N/m

x = Compression of spring = 0.55 m

Here, the kinetic energy of the spring and rocket will balance each other

$\frac{1}{2}mu^2=\frac{1}{2}kx^2\\\Rightarrow u=\sqrt{\frac{kx^2}{m}}\\\Rightarrow u=\sqrt{\frac{800\times 0.55^2}{2}}\\\Rightarrow u=11\ m/s$

The initial velocity of the rocket is 11 m/s = u.

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² = g

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11^2}{2\times -9.81}\\\Rightarrow s=6.16717\ m$

The maximum height of the rocket will be 6.16717 m

Potential energy is given by

$P=mgh\\\Rightarrow P=2\times 9.81\times \frac{0^2-11^2}{2\times -9.81}\\\Rightarrow P=121\ J$

The potential energy of the rocket at the maximum height will be 121 Joules

5 0

4 years ago

A current of 0.86 a flows through a copper wire 0.47 mm in diameter when it is connected to a potential difference of 15 v. how

likoan [24]

To calculate the length of the wire, we use formulas,

$R = \frac{V}{I}$ (A)

$R= \rho \frac{l}{A}$ (B)

Here, R is the resistance of the wire, I is the current flows through wire and V is potential difference. A is cross sectional area of wire and $\rho$ is the density of copper wire and is value, $\rho = 1.7\times 10^{-8} \Omega m$ .

Given $I = 0.86 A,V=15 V and r = \frac{0.47 mm}{2} =2.35 \times 10^{-4} m, V= 15 V.$

Substituting the values of I and V in equation (A ) we get,

$R=\frac{15V}{0.86A} = 17.44 \Omega$

Now from equation (B),

$l=\frac{R A}{\rho }$

Therefore,

$l= \frac{17.44\times\pi \times r^2 }{1.7\times 10^{8} \Omega m} \\\\ l= \frac{17.44\times 3.14 \times(2.35\times10^{-4}m)^2 }{1.7\times 10^{-8} \Omega m} = 177.9 m$

Thus the length of the copper wire is 177.9 m.

8 0

3 years ago