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Nostrana [21]
3 years ago
8

A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,

as shown by the vectors for weight W and normal force N.
a. the net force on the crate is zero or greather than zero. Why?
b. Evidence for this is ____?
Physics
1 answer:
stealth61 [152]3 years ago
4 0

The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it.  That's sad and
disappointing, but I think the question can be answered without seeing
the picture.

The net force on the crate is zero.  Evidence for this is that fact that
the crate is just sitting there.  If the net force on an object is not zero,
then the object is accelerating ... it's either speeding up, slowing down,
or its the direction of its motion is changing.  If none of these things is
happening, then the net force on the object must be zero.

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A spring has a spring constant of 5N/cm. What is its extension when loaded with 15N?
kakasveta [241]
Pretty sure a it’s d since every cm increases by 5N meaning 3 cm would increase it by 15N
3 0
2 years ago
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Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charg
ankoles [38]

Answer:

QA = 19μC

QB = 9.5 μC

Explanation:

  • The force that each charge exerts on the other must obey Coulomb's Law, as follows:

       F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}}  (1)

  • We know that the value of the magnitude of FAB  is 45.0 N, the distance between QA and  QB is 0.19 m, and that QA = 2*QB.
  • Replacing in (1), we can solve for QB, as follows:

      Q_{B} = \sqrt{\frac{F_{AB}*r_{AB} ^{2}}{2*k} } = \sqrt{\frac{45.0N*(0.19m) ^{2}}{2*9e9N*m2/C2} } = 9.5e-6 C  (2)

  • Since QA = 2*QB
  • ⇒ QA = 2* 9.5μC = 19.0 μC
  • ⇒ QB = 9.5μC
5 0
2 years ago
What is the magnification when an object is placed at 2f from the pole of the convex mirror? 
Gekata [30.6K]

Answer:

Linear magnification = 1/3

Explanation:

Given:

Convex mirror

Object's distance from pole = 2f

Find:

Linear magnification

Computation:

Object distance, u = −2f

So,

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/f

1/v = 1/f + 1/2f

BY taking LCM

1/v = 3 / 2f

v = 2f / 3

Magnification, M = -v / u

So,

Magnification, M = (2f / 3) / 2f

Magnification, M = 2f / 6f

Magnification, M = 2 / 6

Linear magnification = 1/3

 

5 0
2 years ago
A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend
solong [7]

Answer:

When her hands extends, her momen of inertia is 4.28\ kg-m^2.

Explanation:

Given that,

Initial angular speed, \omega_i=8\ rad/s

Initial moment of inertia, I_1=100\ kg-m^2

Final angular speed, \omega_f=7\ rad/s

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

I_1\omega_1=I_2\omega_2

I_2 is final moment of inertia

I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2

So, when her hands extends, her momen of inertia is 4.28\ kg-m^2. Hence, this is the required solution.

7 0
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