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grigory [225]
3 years ago
5

Setting up a home network using wireless connections is creating a _____.

Physics
1 answer:
Lemur [1.5K]3 years ago
6 0

A WAN

I just did it, let me know if it was right :)

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a 2 meter tall astronaut standing on mars drops her glasses from her nose. how long will the astronaut have before he hits the g
lorasvet [3.4K]
Here,
Height (S) = 2m
Gravity on mars (g) = 3.7m/s^2
Initial velocity (u) = 0 m/s^2
By the one of the formula of the motion,
S = ut + 1/2at^2
2 = 0 * t + 1/2*3.7*t^2
2 = 1.85t^2
t^2 = 2/1.85 = 1.081
t =1.03s
So, it will take 1.03s long..

7 0
3 years ago
What is a scale on a map?
Oksi-84 [34.3K]
A map is almost always smaller than the place it describes. If a map of the US were drawn in its actual size, it would be 3,000 miles wide, and very difficult to fold. ... The scale of the map is the ratio of a distance on the map to the same distance on the real thing. ... If the map scale is 1 : 50000, then 1 foot on the map shows things that are actually spread over 50000 feet in the real city or field.
7 0
3 years ago
An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof
Pavel [41]

Answer:

T = 69^o C

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

Q_{ice} = Q_{coffee}

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2

here we have

m_1 = 15 gram

L = 333 kJ/kg = 333 J/g[/tex]

c_1 = c_2 = 4186 J/kg C = 4.186 J/g C

\Delta T_1 = T - 0

\Delta T_2 = 85 - T

now we have

15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)

4995 + 62.79T = 49813.4 - 586.04T

648.83 T = 44818.4

T = 69^o C

6 0
3 years ago
Can someone help please and thank you:)
Ray Of Light [21]

Answer:

The answer is C.

Explanation:

4 0
2 years ago
A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
3 years ago
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