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3 years ago

The Sun transfers energy to the atmosphere primarily by _____.

1 answer:

5 0

Radiation is the only one that can pass through a vaccum (like space). You need a medium for conduction and convection to work. Therefore the answer is radiation.

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A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

podryga [215]

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

3 0

3 years ago

Read 2 more answers

The table below lists the speed that sound travels through four different materials.One of these materials is a gas, two are liq

Firdavs [7]

In order to find the solid, you would want the object in which sound travels the fastest

In this case, since in object C, the speed of sound is the fastest, it is the most likely to be a solid

So object C is most likely to be a solid

5 0

1 year ago

A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with

mariarad [96]

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

$F_friction=u_s*N=u_s*(mg)$

$F-F_friction=ma$

$F-u_s*(mg)=ma$

Second Newton law for the Big Block:

$F_friction=7ma$

$u_s*(mg)=7ma$

$a=u_s*g/7$

8 0

3 years ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0

3 years ago

In a thundercloud there may be electric charges of 45.0 C near the top of the cloud and -45.0 C near the bottom of the cloud. Th

Gala2k [10]

Answer:

The electric force on the top charge is $F=3.44\times 10^6\ N$ .

Explanation:

Given that,

Electric charges in a thundercloud, $q_1=q_2=45\ C$

The distance between charges, d = 2.3 km = 2300 m

Let F is the electric force on the top charge. The electric force is given by the formula as :

$F=\dfrac{kq^2}{d^2}$

$F=\dfrac{9\times 10^9\times (45)^2}{(2300)^2}$

$F=3445179.58\ N$

$F=3.44\times 10^6\ N$

So, the electric force on the top charge is $F=3.44\times 10^6\ N$ . Hence, this is the required solution.

7 0

3 years ago