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Molodets [167]
3 years ago
13

Mary's friend observes the investigation and claims that block B has more mass. If there were no errors in the investigation, wh

ich of the following must be TRUE for this claim to be correct?
Physics
2 answers:
Elena L [17]3 years ago
6 0
This is a true claim.
Keith_Richards [23]3 years ago
3 0

Answer:

Block B must have greater density than block A.

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The system needs an ordinary friction-based brake to bring the train to a full stop. Explain why the magnetic brake is not very
BabaBlast [244]

Answer:

The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.

Explanation:

A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.

Let us find the drag force which is proportional to the velocity of two conducting plates.

The EMF that is induced in the eddy currents are : $E=v(B \times L)$

The force which is due to the induced magnetic field is, $F=l(L \times B)$

Therefore, $F=\frac{E}{R} \times (L \times B)$

                 $F=\frac{v(B \times L)}{R} \times (L \times B)$

Here, force is directly proportional to the velocity of the two conducting plates.

Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.  

6 0
3 years ago
Molecular homologies include similarities between all of the following except ___________.
dezoksy [38]

Answer:

Punctuated sedimentation

Explanation:

Punctuated sedimentation isn't part of the molecular homologies or the theory of molecules

6 0
3 years ago
You are in a contest with your friends to see who can drive a golf ball the farthest. should you hit a "line drive" (low to the
WITCHER [35]
Line drive would be more farthest from the very high angle shot because when we increase the angle of flight the range started to increase and at some point it becomes maximum and that angle is 45° after that as you go on increasing the angle it won't be covering more distance as compared to the max at (45°) .
3 0
3 years ago
If the car went 30 km west in 25 min. and then 40 km east in 35 min., what would be its total displacement?
ozzi
Displacement only contain the length of movement and the direction of that movement
So you can pretty much disregard the speed.

30 to the west than 40 to the east

The displacement will be 10 km to the east

hope this helps
8 0
3 years ago
Read 2 more answers
A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them
Tatiana [17]

Answer:

A)    q₂ = 75.98 cm, B)     q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

          \frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}

         \frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}  

          1 / q₁ = 0.04756

           q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

          p₂ = 39.4 -q₁

          p₂ = 39.4 -21.0227

          p₂ = 18.38 cm

let's use the constructor equation

            1 / q₂ = 1 / f - 1 / p2

             

             \frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}

            \frac{1}{q_2} = 0.01316

            q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

            q₂’= q₂ + 39.4

             q₂'= 75.98 +39.4

              q₂' = 115.38 cm

C) the magnification of a lens is

              m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

          m = - 75.98 / 50

          m = -1.5 X

the negative sign indicates that the image is inverted

4 0
3 years ago
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