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kompoz [17]
3 years ago
9

Which type of star is the sun

Physics
1 answer:
ahrayia [7]3 years ago
3 0

G2V

The sun is a G2v type of star, a yellow dwarf and main sequence star.

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Which symbol and unit of measurement are used for electric current?
Burka [1]

Answer: Symbol is I and unit A

Explanation: A represents Amperes

HOPE THIS HELPS!!!!!!!!

4 0
3 years ago
How do fossils provide evidence of continental drift?
Alex777 [14]
C explanation: There are many examples of fossils found on separate continents and nowhere else, suggesting the continents were once joined. If Continental Drift had not occurred, the alternative explanations would be: They swam to the other continent/s in breeding pairs to establish a second population. ...
8 0
3 years ago
Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want
denpristay [2]

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of  the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

4 0
3 years ago
Read 2 more answers
While on a moving elevator during a certain perfod or time, Frank's apparent weight is 620 N. If Frank's mass is 70 kg, what is
MakcuM [25]

Answer:

0.94 m/s^2 downwards

Explanation:

m = 70 kg, m g = 70 x 9.8 = 686 N

R = 620 N

Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.

mg - R = ma

686 - 620 = 70 x a

a = 0.94 m/s^2

7 0
3 years ago
An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror
andrew-mc [135]

Answer:

Position = \frac{60}{7}\ cm behind the mirror

Nature = Virtual and Erect

Size = \frac{15}{7}\ cm : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = =\frac{h_{image}}{h_{object}}=-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}

Height of the object = 5 cm

Height of the image = 5\times\frac{3}{7}=\frac{15}{7}\ cm

Since the height of the image is positive and less than the size of object,it is erect and diminished.

4 0
3 years ago
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