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Alchen [17]
3 years ago
7

What is the weight on earth of a girl with a mass of 17 kg?

Physics
1 answer:
madreJ [45]3 years ago
8 0
The gravity on earth is about 9.8 thus you times the mass by the gravity on earth:
17*9.8 = 166.6

Hope this helps! :)
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I need help for a assignment NOT A QUIZ
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Answer:

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3 years ago
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Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?
oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

6 0
3 years ago
Problems - Show all work.
Fittoniya [83]

Answer:

21s

Explanation:

Given parameters;

Radius  = 10m

Speed or velocity  = 3m/s

Unknown:

Period  = ?

Solution:

To solve this problem, use the expression:

      v  = \frac{2\pi r}{T}  

r is the radius

T is the unknown

           Input the parameters and solve for T;

    3  = \frac{2 x \pi  x 10}{T}  

    62.84 = 3T

         T  = 21s

4 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?
pishuonlain [190]

The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net       - ( 1 )

Step 1:

The net electric field value is given as:

E_net  = E₁ cos Φ + E₂ cos Φ      

           = 2E₁ cos Φ                  -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod                

            respectively.

            Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r)             - ( 3 )

where, k is Coulomb's force constant

            λ is linear charge density

            r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

           a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

  = √1.08

  = 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

   = k [(Q/x)/r]

   = k [ Q/xr ]

   = (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

   = 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:    

E₂ = k [ Q/xr ]

    = (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

    = 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net  = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

   = tan⁻¹ ( 1 )

   = π/4  

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

          = 2(1.803×10⁴ N/C) (0.5)

          = 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

  = 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is  2.885 × 10⁻¹⁵ N.

Learn more about electric force here:

<u>brainly.com/question/1634182</u>

#SPJ4      

8 0
2 years ago
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