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Elodia [21]
3 years ago
15

If someone drops a cup, it falls to the ground. Why doesn't the gravitational force between the person's hand and the cup keep t

he cup from falling?
Physics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

Explanation:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
Find the value of 02?
TiliK225 [7]
My educated guess : 21.2 deg
3 0
3 years ago
Plz answer this very soon
tester [92]

Answer:

Im gonna say it is answer A:) Hope this helps!

Explanation:

6 0
3 years ago
Read 2 more answers
Which two elements define a story's setting?
Archy [21]

Answer:

the plot structure defines a story's setting

6 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.5 km/h and the μs
Yakvenalex [24]

Answer:

Radius=15.773 m

Explanation:

Given data

v=29.5 km/h=8.2 m/s

μs=0.435

To find

Radius R

Solution

The acceleration is a centripetal acceleration  which is experienced by the bicycle given by

a=v^{2}/R

This acceleration is only due to static force which given as

f=ma\\f=m(v^{2}/R )

The maximum value of the static force is given as

Fs_{max}=u_{s}F_{N}

where

FN is normal force equal to mass*gravity

Therefore when the car is on the verge of sliding

f=fs_{max}\\ m(v^{2}/R )=u_{s}mg

Therefore the minimum radius should be found by the bicycle move without  sliding

So

v^{2}/R=u_{s}g\\  R=v^{2}/u_{s}g\\R=(8.2)^{2}/(0.435*9.8)\\R=15.773m

8 0
3 years ago
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