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Otrada [13]
3 years ago
8

When do websites collect personal and private information?

Physics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

when an acount is made online i can also give you the other awnsers

Explanation:

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Calculate the kinetic energy of Earth due to its spinning about its axis. Krot = J Compare your answer with the kinetic energy o
Studentka2010 [4]

Answer:

Explanation:

Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.

Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.

Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.

Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS

8 0
4 years ago
John runs for 10 mins at a uniform speed 9kn/h. At what speed should he run for the next 20mins so that theaverage speed comes t
kolbaska11 [484]

Answer:

13.5 km/h

Explanation:

John runs for 10 min at 9 km/h, then 20 min at v.  He will have run a total of 30 min at 12 km/h.

Therefore:

(30 min) (12 km/h) = (10 min) (9 km/h) + (20 min) v

v = 13.5 km/h

4 0
3 years ago
For an object that is speeding up at a constant rate how would the acceleration vs. time graph look?
frosja888 [35]

Answer:

C

Explanation:

Horizontal is constant... if it was on x axis it would be a speed of 0

Please give brainliest answer

5 0
3 years ago
Read 2 more answers
What affects the braking distance and thinking distance
Radda [10]
Your speed is one of the only factors that has an effect on both your thinking distance and braking distance. Put simply, the faster you are going, the greater the distance travelled before you apply the brakes (thinking distance) and the vehicle comes to a complete stop (braking distance).
8 0
4 years ago
Read 2 more answers
An object of mass m is dropped from height h above a planet of mass M and radius R .
Margarita [4]

Answer:

v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}

Explanation:

The initial mechanical energy of the object, when it is located at height h above the the planet, is just gravitational potential energy:

E=U=\frac{GMm}{(R+h)}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

h is the altitude of the object

When the object hits the ground, its mechanical energy will sum of potential energy and kinetic energy:

E=\frac{GMm}{R}+\frac{1}{2}mv^2

where

v is the speed of the object at the ground

Since the mechanical energy is conserved, we can write

\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}

and solving for v, we find

\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}\\v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}

3 0
3 years ago
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