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ANTONII [103]
3 years ago
5

How can I skip class:

Physics
2 answers:
Dahasolnce [82]3 years ago
3 0

Answer:

c gl

Explanation:

::::::::::::::::::::::::

vekshin13 years ago
3 0
Thanks for the free points I would do C
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What happens when electromagnetic waves cause a disturbance in electric
Leviafan [203]

Answer:

c

Explanation:

just answered it

7 0
2 years ago
Which part of a river would have animals with very muscular bodies and adaptations that let survive in turbulent water? Source z
lidiya [134]
The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise. Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here. Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.
8 0
3 years ago
Read 2 more answers
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
3 years ago
-A pool ball leaves a 0.70 meter high table with an initial velocity of 2.4 m/s. Predict the time required for the pool ball to
lorasvet [3.4K]

Answer:

0.70+2.4=23.7+35.0=answer

Explanation:

4 0
3 years ago
Read 2 more answers
In designing a backyard water fountain, a gardener wants to stream of water to exit from the bottom of one tub and land in a sec
mote1985 [20]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.

From the definition we know that the speed of a body can be described as a function of gravity and height

V = \sqrt{2gh}

V = \sqrt{2*9.8*0.15}

V = 1.714m/s

Then applying the kinematic equation of displacement, the height can be written as

H = \frac{1}{2}gt^2

Re-arrange to find t,

t = \sqrt{2\frac{h}{g}}

t = \sqrt{2\frac{0.5}{9.8}}

t = 0.3194s

Thus the calculation of the displacement would be subject to

x = vt

x =1.714*0.3194

x = 0.547m

Therefore the required distance must be 0.547m

4 0
3 years ago
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