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tia_tia [17]
2 years ago
11

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

eased from rest at a vertical position that is not the system’s equilibrium position. The system oscillates near Earth’s surface. The system is then taken to the Moon’s surface, where the gravitational field strength is nearly 16 that of the gravitational field strength near Earth's surface. Which of the following claims is correct about the period of oscillation for the system?
Physics
1 answer:
vodka [1.7K]2 years ago
4 0

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

T= 2\pi\sqrt{\frac{M}{k}}

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

mg = kx

x = \frac{mg}{k}

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

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Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
Veronika [31]

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

3 0
2 years ago
Suppose that, instead of the Coulomb force law, one finds experimentally that the force between any two charge q1 and q2 is Writ
denpristay [2]

Answer: E= KQ/r^2

Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.

The magnitude of an electric field E, at a point is given by Coulombs law as

E/ F/q

Where F= Coulombs force exertedon the charge and q= electric charge

E= F/q=(KQq)/r^2q

E=KQ/r^2

6 0
2 years ago
A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th
agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

T - mg = ma

here we know that maximum possible acceleration of so that string will not break is given as

T = 54 N

now we have

54 - (5 \times 9.8) = 5 a

a = 1 m/s^2

now for such acceleration we can use kinematics

d = \frac{1}{2}at^2

2 = \frac{1}{2}(1) t^2

t = 2 s

7 0
2 years ago
The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect
Diano4ka-milaya [45]
Rx= 3.5 km

Ry= 2.9 km
4 0
2 years ago
Read 2 more answers
Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of
gtnhenbr [62]

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

5 0
2 years ago
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