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3 years ago

Witch scientific tool helps to make distant objects nearer and larger?

Physics

1 answer:

vladimir2022 [97]3 years ago

8 0

Answer:

magnifying glass

Explanation:

makes objects bigger and smaller / used in science

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Vector A has a magnitude of 4.0m and points 30 south of east vector B has a magnitude of 2.0m and points 30 north of west the re

valentina_108 [34]

Answer:

2.06 m at 49.7 or 50 degree South of East

Explanation:

take x axis and y axis and follow as explained in attachment

Download pdf

6 0

3 years ago

sineoko [7]

Answer:

a) λ = 5,084 10⁻⁷ m , b) P = 3.63 10²⁶ W , c) P = 5.8 10²⁷ W and d) λ = 2.54 10⁻⁷ m

Explanation:

a) The maximum emission of the sun can be calculated using the Win equation

λ T = 2,898 10⁻³ m.K

λ = 2,898 10⁻³ / T

λ = 2,898 10⁻³ / 5700

λ = 5,084 10⁻⁷ m

λ = 5,084 10⁻⁷ m (1 10⁹ nm / 1m) =

λ = 5,084 10² nm = 508.4 nm

photon in the visible range

b) The emission of the Sun, is described by the Stefan equation

P = σ A e T⁴

Where σ is the Stefan-Boltzmann constant that vslue is 5,670 10-8 W/m²K⁴, A area of the Sun, and e the emissivity that for a perfect black body is 1

In order to use this equation, we must calculate the area of the sun, we consider it a perfect sphere

r = 695,000 km (1000m / 1 km) = 6.95 10⁸ m

Area of a sphere

A = 4π R²

A = 4π (6.95 10⁸8)²

A = 6.07 10¹⁸ m²

P = 5,670 10⁻⁸ 6.07 10¹⁸ 1 5700⁴

P = 3.63 10²⁶ W

c) The new temperature is double the previous one

T = 2 To

Let's substitute in the formula and calculate

P = σ A e (2To)⁴

P = σ A e T⁴ 2⁴

Po = σ A e T4 = 3.63 10 26 W

P = 16 Po

P= 16 (3.63 10²⁶)

P = 5.8 10²⁷ W

d) Let's calculate the explicit value of the temperature and use the Win equation

T = 2 5700

T = 11400K

λ = 2,898 10⁻³ / 11400

λ = 2.54 10⁻⁷ m

λ = 2.54 10²nm = 254 nm

photon in the UV range

5 0

3 years ago

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc

Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

$P=VI$

Express this in terms of 'I'. This gives,

$I=\frac{P}{V}$

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, $P_1=16\ W, V=110\ V$

So, current drawn is given as:

$I_1=\frac{16\ W}{110\ V}=0.145\ A$

For incandescent bulb, $P_2=100\ W, V=110\ V$

So, current drawn is given as:

$I_2=\frac{100\ W}{110\ V}=0.909\ A$

For CFL bulb, $P_3=24\ W, V=110\ V$

So, current drawn is given as:

$I_3=\frac{24\ W}{110\ V}=0.218\ A$

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0

4 years ago

A tennis ball is dropped from 1.43 m above the

Rudiy27

Answer:

-5.29 m/s

Explanation:

Given:

y₀ = 1.43 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)

v = -5.29 m/s

4 0

4 years ago

Average velocity is different than average speed because calculating average velocity involves ?

VashaNatasha [74]

direction ................................................................

6 0

4 years ago

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