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scZoUnD [109]
1 year ago
11

Consider the following FBDs.FBDsAn elevator is suspended by a cable and moves uniformly upward. Which of the above free body dia

grams best represents the forces acting on the elevator?

Physics
1 answer:
Drupady [299]1 year ago
6 0

The force of gravity F_g will act downwards.

Normal force F_N will act upwards equal to the force of gravity.

A force due to uniform acceleration F_a will act upwards to move the elevator upwards.

Thus, figure E is the correct answer.

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Are we actually touching nothing but electrons in reality or can we actually feel things without the electron barier?
Anna007 [38]

Objects in contact with each other never get past the electron clouds of the atoms on their surfaces.

5 0
3 years ago
Two objects were lifted by a machine.One object had a mass of 2 kilograms and was lifted at a speed of 2 m/sec.The other had a m
NikAS [45]

Answer:

While lifting two object the machine needs the different momentum for different mass object.

Explanation:

  • Momentum is the quantity of motion contained in an object. Usually it is measured by the product of mass and velocity.
  • Momentum of first mass = 2 kg × 2 m/sec = 4 kg m/sec
  • Momentum of second mass = 4 kg × 3 m/sec = 12 kg m/sec
  • So the machine requires higher mass in motion for second object ( i.e. momentum) than the first one while lifting.
7 0
3 years ago
The angle of reflection is the angle between the normal line and the___?
Aleks [24]

Answer:

Reflected line

Explanation:

The angle between the reflected ray and the normal ray is called angle of reflection.

4 0
3 years ago
A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring
Taya2010 [7]

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

x = \frac{-F}{K}

Now  

x = \frac{-3500}{14000} \\\\

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

4 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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