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Maslowich
3 years ago
11

To lift an object, you must pull upward on the object with a force __________ to the object’s weight.

Physics
1 answer:
Rudiy273 years ago
7 0

Answer:

To lift an object, you must pull upward on the object with a force greater than the object’s weight.

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The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in
g100num [7]

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

  • If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

  • If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

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3 years ago
Which forces always push in the opposite direction of motion slow stuff down.
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Friction- the external force that acts on objects and causes them to slow down when no other external force acts upon them.
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4 years ago
Read 2 more answers
A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf 2.5m off the ground. How much gravitati
Damm [24]
Gravitational potential energy  = 

                        (mass)  x  (gravity)  x  (height)

                =    (5.8 kg)  x  (9.8 m/s²)  x  (2.5 m)

                =           142.1 Joules         (C)
5 0
3 years ago
What does vf stand for<br> a.fringe velocity<br> b.first velocity<br> c.final velocity
Elza [17]
The correct answer is C. Final Velocity

Hope this helped!
5 0
3 years ago
Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

3 0
3 years ago
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