Round to the hundredths place.

NASA developed a reflective foil barrier to help shield spacecraft from heat transfer in space. These reflective barriers are no

morpeh [17]

Answer: Radiation

Explanation: Radiation is the energy that comes from a source in form of electromagnetic waves, subatomic particles, light, or heat which travels through space.

Examples of radiation include the light, heat, and particles emitted from the Sun.

Using a foil barrier to prevent heat transfer is possible because foil has a silver color, and silver reflects light and heat instead of absorbing them. This is the opposite of black surfaces that absorb heat.

So in homes where these foil reflective barriers are used, the transfer of heat through Radiation is highly reduced.

3 0

3 years ago

If two balls have the same volume,

Lena [83]

Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

1. Ball A will have the greater density
2. Ball C and Ball D have the same density.
3. Ball Q will have the greater density.
4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

Density = Mass/Volume.

For Case 1:

Va = Vb and Ma = 2Mb
D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
Therefore, the density of ball A,
D(a) = 2D(b).
Therefore, ball A has the greater density.

For Case 2:

Vc = 3Vd,

Vd = (1/3)Vc

Md = (1/3)Mc

D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

D(c) = D(d).

Therefore, ball C and D have the same density

For Case 3:

Vp = 2Vq and Mp = Mq
D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
Therefore, the density of ball P is half the density of ball Q
Therefore, ball Q has the greater density.

For case 4:

Mx = (1/2)My
Vx = Vy

Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

8 0

2 years ago

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

2 years ago

A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe, the velocity is given by

$v=\frac {2Lf}{n}$

Making L the subject then

$L=\frac {nV}{2f}$

Where f is the frequency, L is the length, n is harmonic number, v is velocity

Substituting 1 for n, 320 Hz for f and 331 m/s for v then

$L=\frac {1*331}{2*320}=0.5171875\approx 0.52m$

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, $f2=2\times \frac {v}{2L}$ and $f3=3\times \frac {v}{2L}$

$f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz$

(c)

When v=367 m/s then

$f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz$

5 0

3 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

3 years ago