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irga5000 [103]
3 years ago
10

Planet B has a tilt of 45 degrees. What seasonal changes would be expected on this planet?

Physics
1 answer:
12345 [234]3 years ago
5 0
... Extreme temperature changes between seasons.

... Extreme changes in length of daylight and darkness.

... Larger areas around both poles where daylight/darkness
can last more than a whole day. 
Altogether, these areas cover half of the planet. 
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A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0
3 years ago
Question 7 of 10
Dahasolnce [82]
Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion.
https://www.khanacademy.org › stat...
Static and kinetic friction example (video) | Khan Academy

Answer a would be static friction
Answer b is fluid friction
(Air resistance is fluid friction. Fluid friction is the friction experienced by objects which are moving in a fluid and the air is a fluid.)
Answer c is static friction
ANSWER D IS KINETIC FRICTION

Hope this helps :D



4 0
2 years ago
HELP!!!! WILL MARK BRAINLIEST!!!! IF YOU LEAVE AN ANSWER EXPLAIN! THANKS
Travka [436]

Answer:

the answer is C

Explanation:

we know this because if you compare the graphs and look at the direction. it isn't always in the explanation or the few sentences they gave you at the top. also, look at the waves, you can see in Davids drawing that it is directly straight up, A and B do not represent that. A isn't even a valid answer. Notice also in A that the arrow is going in the completely different direction than in Davids drawing. B is also going a different direction even though it is only turned a little bit although if it was straight up like Davids drawing then it would most likely be a correct answer. C does have one arrow going a different direction but look at how it has two, showing in which if the waves were to turn then the arrow is still valid

7 0
2 years ago
A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².​
Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2}  } =0.0001[m^{2} ]

Now replacing:

P=100/0.0001\\P=1000000[Pa]

3 0
3 years ago
A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter
icang [17]

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

7 0
2 years ago
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