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Vedmedyk [2.9K]
1 year ago
8

Which statement accurately describes the motion of the object in the graph above over 10 seconds? Group of answer choices The ob

ject is at rest for 3 seconds and then decreases velocity to 1 cm/s for 3 seconds. The object is at rest for 4 seconds and then moves forward for 4 seconds at 1 cm/s. The object moves forward at 1 cm/s, stops, and then increases velocity to 2 cm/s. The object moves forward at 1 cm/s, stops, and then continues at the same velocity.

Physics
1 answer:
timurjin [86]1 year ago
7 0

Answer:

Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec     (11 - 3) / (10 - 6) = 2 cm/sec

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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through 90^{o} then position of two mirrors will be changed. The resultant path difference will be as follows.

      \frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}

Formula for change in fringe shift is as follows.

          n = \frac{2lv^{2}}{\lambda c^{2}}

       v^{2} = \frac{n \lambda c^{2}}{2l}

             v = \sqrt{\frac{n \lambda c^{2}}{2l}}

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

             l = 11 m

    \lambda = 5.9 \times 10^{-7} m

           c = 3.0 \times 10^{8} m/s

Hence, putting the given values into the above formula as follows.

            v = \sqrt{\frac{n \lambda c^{2}}{2l}}

               = \sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}

               = 2.41363 \times 10^{9} m/s

Thus, we can conclude that velocity deduced is 2.41363 \times 10^{9} m/s.

3 0
2 years ago
An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo
lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

\frac{d'_1}{d'_2}=\frac{f_1}{f_2}

Magnification of first lens

M_2=-\frac{d'_1}{d_1}

                   and

M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

                   and

\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

which means,

\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

and

\frac{d'_1}{d_2'}=\frac{f_1}{f_2}

So, we get

\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm

∴ Focal length should this camera's lens is 24.71 mm

6 0
2 years ago
Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s.
Nataliya [291]

Answer:

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7 0
1 year ago
Compute the mean and maximum velocities for a liquid with a flow rate of 20 L/min in a 1.5-in nominal diameter sanitary pipeline
vlada-n [284]

Answer:

Mean velocity = 0.292 m/s

Maximum velocity = 0.584 m/s

The flow is laminar as Re = 229.2

Explanation:

D = 1.5 inches = 0.0381 m

Q = volumetric flow rate = 20 L/min = 0.000333 m³/s

Q = A × v

A = Cross sectional Area = πD²/4 = π(0.0381)²/4 = 0.00114 m²

v = average velocity

v = Q/A = 0.000333/0.00114 = 0.292 m/s

For flow in circular pipes, maximum velocity = 2 × average velocity = 2 × 0.292 = 0.584 m/s

To check if the flow is laminar or turbulent, we need its Reynolds number

Re = (ρvD)/μ

ρ = 1030 kg/m

v = 0.292 m/s

D = 1.5 inches = 0.0381 m

μ = 50 cP = 0.5 poise = 0.05 Pa.s

Re = (1030 × 0.292 × 0.0381)/0.05 = 229.2

For laminar flow, Re < 2100

For turbulent flow, Re > 4000

And 229.2 < 2100, hence, this flow is laminar.

7 0
2 years ago
A typical mattress has a network of springs that provide support. If you sit on a mattress, the springs compress. A heavier pers
GenaCL600 [577]

Answer:

Explanation:

Spring has a tendency to store energy in them and deform its shape when force is applied on it. Once the applied force is removed it regains its original shape and size.

It is in helical shape and is used in mattress to give structure and support. Spring have elastic nature and follows spring forces, F = k * x

where is the applied force, k is the spring constant and x is the amount of extension.

When a heavier person sits on a mattress, more weight is applied on springs and they form coils, as weight is removed they regains its shape again.

4 0
2 years ago
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