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DochEvi [55]
3 years ago
12

Someone help me please

Physics
2 answers:
riadik2000 [5.3K]3 years ago
6 0

the answer is 25 newtons

aleksley [76]3 years ago
5 0
Each time the distance doubles the gravity quarters, then at two earth radii it will weigh 25 Newton’s
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Answer:

7200N

Explanation:

Centripetal force is directly proportional to the product of the mass and the square of the velocity and inversely proportional to the radius given.

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What is the unit of work in SI system?
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Answer:

Joule ;)

Explanation:

In the case of work (and also energy), the standard metric unit is the Joule (abbreviated J). One Joule is equivalent to one Newton of force causing a displacement of one meter. In other words, The Joule is the unit of work.

Hope this helps!

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3 years ago
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Which statement is true regarding flexibility? A. joint's range of motion cannot be maintained with age. b.A joint's range of mo
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3 years ago
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During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.

4 0
3 years ago
At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh
Anna35 [415]

We have the equation for electric field E = kQ/d^{2}

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to d^{2}

So, \frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}

E_{1} = 485 N/C

d_{1} = 0.208 cm

d_{2} = 0.620 cm

E_{2} = ?

\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}

E_{2} = \frac{485*0.208^{2}}{0.628^{2}}

E_{2} = 53.20 N/C

7 0
3 years ago
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