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Lelu [443]
3 years ago
15

HELP ME PLEASEEEEEEEEEEEEEE

Physics
1 answer:
rjkz [21]3 years ago
7 0

Answer:

The one you have selected is correct.  :)

Explanation:

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The angular separation of two stars is 0.1 arcseconds and you photograph them with a telescope that has an angular resolution of
gizmo_the_mogwai [7]

Answer:

Will see them as only one star

Explanation:

Solution:

- The angular resolution of a telescope means the minimum quantity that can be visualized. Since their angular separation ( 0.1 arcseconds ) is smaller than the telescope's angular resolution (1 arcseconds ), your photograph will seem to show only one star rather than two.

7 0
3 years ago
Which statement is TRUE?
Reptile [31]

Answer:

energy ina battery makes new energy called electrical energy

7 0
3 years ago
A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a
IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}

Put the value into the formula

\dfrac{dM}{dt}=53.3+3.63

\dfrac{dM}{dt}=56.93\ kg/s

We need to calculate the thrust of the jet engine

Using formula of thrust

T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v

Put the value into the formula

T=56.93\times317-53.3\times260

T=4188.81\ N

Hence, The thrust of the jet engine is 4188.81 N.

7 0
4 years ago
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​
poizon [28]

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

6 0
3 years ago
A 50 mm diameter thin walled pipe is covered with an insulation layer with thicknessof 25mm and thermal conductivity of0.075W/mK
Wittaler [7]

Answer:

The steam will start to condense at 6.6 mm into the pipe

Explanation:

The volume flow rate =π×(50/1000)²/4×10 = 0.0196 m³/s

The specific volume of the steam = 1.769 m³/kg

Therefore;

The mass flow rate = 0.0196/1.769 = 0.011099  kg/s

The resistance of the insulation material = ln(0.075/0.05)/(2×π×0.075) = 0.860 K/W

The resistance of the outside film of the insulator = 1/(15×2×π×0.075×1) = 0.14147 K/W

The total resistance = 0.14147 + 0.860 = 1.00147 K/W

1/(UA) = 1.00147 K/W

A = 2×π×0.05×1

1/U = 0.3146

U = 3.178 W/m² K

We have;

T(x) = T₀ + (Tin - T₀) exp(-UπDx/mcp)

Therefore, when T(x) = 100°C, we have;

100 = 20 + (120 - 20)exp(-3.178×π×0.05x/(0.011099 × 1.33))

Solving, we get

x = 6.597× 10⁻³ m ≈ 6.6 mm

Therefore, the steam will start to condense at 10 mm into the pipe.

3 0
4 years ago
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