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Rzqust [24]
3 years ago
7

Why does our sun appear larger than any other star in our galaxy

Physics
2 answers:
MArishka [77]3 years ago
5 0

Answer:

It is the closest star to us. other bigger stars may seem small, but that is because they are far away. The more closer an object is, the smaller it seems to be.

Explanation:

Hope this helps!

Doss [256]3 years ago
3 0

Answer:

The closer and bigger the star, the larger the star appears.

Explanation:

Since the sun is closer to us it shows up way larger than other stars in our galaxy.

Have a great day.

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20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0
3 years ago
Which is a way to improve self esteem?
erik [133]

Answer: Positive Talk

Explanation:

5 0
3 years ago
Hello, why do all electromagnetic waves travel at the same speed in vacuum?
Viktor [21]

Answer:

due to the magnetic field

Explanation:

magnetic field is the same in the vacuum

6 0
3 years ago
A 50-ohm resistor in a 0.5 Amp circuit produces how much voltage drop?​
ahrayia [7]

25 volts

Explanation:

Use Ohm's law to find the potential drop:

V = IR

= (0.5 A)(50 ohms)

= 25 volts

6 0
3 years ago
A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high
Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}

Where:

v_{o} - Initial velocity, measured in meters per second.

y - Final height, measured in meters.

y_{o} - Initial height, measured in meters.

t - Time, measured in seconds.

g - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}

Given that y = 0\,m, v_{o} = 15\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and t = 16\,s, the final height of the package is:

y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}

y_{o} = 1015.296\,m

The package was released at a height of 1015.296 meters.

7 0
3 years ago
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