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Anna35 [415]
3 years ago
13

Twovectors Å and °B are such that A = 2and A-B-C find the angle between them​

Physics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

hjg567 A 567

Explanation:

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falling objects drop with an average acceleration of 9.8m/sec/sec. if an object falls from a tall building how long will it take
sveta [45]

Answer:

5 seconds

Explanation:

<em>Acceleration = (final velocity - initial velocity) ÷ time</em>

<em>a =  \frac{v - u}{t}</em>

<em>9.8 =  \frac{49 - 0}{t}</em>

<em>9.8 =  \frac{49}{t}</em>

<em>9.8t = 49</em>

<em>t =  \frac{49}{9.8}</em>

<em>t = 5</em>

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3 years ago
Charges that do not transfer
ratelena [41]

Answer:what is the question exactly

Explanation:

8 0
3 years ago
Why can work be thought of as the transfer of energy? Please help it's physics
deff fn [24]
That is because work requires energy. According to the law of conservation of energy, it cannot be created or destroyed. When doing work, energy change forms and gets transferred to the object until it is released.
for example, when you lift up an object and place it on a higher elevation, you transferred energy to it and gave it potential energy. The potential energy is transformed into kinetic energy when the object falls down, and if it hits a surface, the energy will scatter, vibrating the areas around it and producing sound.
Also, work= force X distance. The energy does not go away, but rather get changed into some other form of energy
7 0
3 years ago
Read 2 more answers
When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr
Scrat [10]

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

7 0
3 years ago
An athlete with mass m running at speed v grabs a light rope that hangs from a ceiling of height H and swings to a maximum heigh
-Dominant- [34]

Answer:

d) The 2 athletes reach the same height, because the athletes run with the same speed.

Explanation:

In the whole process , kinetic energy is converted into potential energy .

1/2 m v² = mgh

v² = 2gh

h = v² / 2g

In this expression we see that height attained does not depend upon mass of the object . At the same time it also makes it clear that it depends upon velocity . As the velocity in both the cases are same , height attained by both of them will be same. Hence option d ) is correct.

7 0
3 years ago
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