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solniwko [45]
4 years ago
8

The distance versus this plot for a particular object shows a quadratic relationship. Which column of distance data is possible

for this solution?​

Physics
1 answer:
SOVA2 [1]4 years ago
5 0

Answer:

A

Explanation:

In Column A, the pattern is 1², 2², 3², etc.  So x(t) = t², which is quadratic.

In Column B, the pattern is 2*1, 2*2, 2*3, etc.  So x(t) = 2t, which is linear.

In Column C, the pattern is 9*1, 9*2, 9*3, etc.  So x(t) = 9t, which is linear.

In Column D, the pattern is 1/1, 1/2, 1/3, etc.  So x(t) = 1/t, which is reciprocal.

In Column E, the pattern is 1/1², 1/2², 1/3², etc.  So x(t) = 1/t², which is reciprocal.

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A current of 4.50 a flows through a resistor that has r = 6.20 ω. what is the voltage across the resistor, that is, the potentia
Natali5045456 [20]
I = 4.50 amps
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4 0
3 years ago
Tendons are, essentially, elastic cords stretched between two fixed ends; as such, they can support standing waves. These resona
Hatshy [7]

Answer:

161.938 Hz

Explanation:

the computation of the fundamental resonant frequency is shown below

p = 1100 kg/m^3

A = 130 mm^2

= 130 ×10^-6 m^2

T = 600 N

L = 20 cm

= 0.2 m

Now the linear density of tendon is

=  1100 kg/m^3 ×  130 ×10^-6

= 0.143 kg/m

Now the wave of the string is

= √600  ÷ √0.143

= 64.775 m/s

Now finally the fundamental resonant frequency is

= 64.775 ÷ (2 × 0.2)

=161.938 Hz

3 0
3 years ago
In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35 degree angle compressing the spring 18 cm ca
Free_Kalibri [48]

Ball hits the floor at 1.6 m below the initial position

So here we can say that

\Delta y = -1.6 m

also it will hit at horizontal distance of 5.2 m

\Delta x = 5.2 m

let say its velocity in x and y directions are given as

v_x ,v_y

now we can say

v_x* t = 5.2

-1.6 = v_y * t - \frac{1}{2}gt^2

from above two equations

-1.6 = v_y* \frac{5.2}{v_x} - 4.9t^2

as we know that it is projected at an angle of 35 degree

so we know that

v_y = v_x tan35

-1.6 = 5.2 tan35 - 4.9 t^2

4.9 t^2 = 3.64 + 1.6 = 5.24

t = 1.03 s

now we have

v_x * 1.03 = 5.2

v_x = 5.03 m/s

also we can find vertical component of velocity as

v_y = v_x tan35 = 3.52 m/s

now we will find net velocity as

v^2 = v_x^2 + v_y^2

v^2 = 5.03^2 + 3.52^2

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now by energy conservation we can say

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

k*(0.18)^2= 0.032*(6.14)^2

k = 37.2 N/m

so spring constant is 37.2 N/m

5 0
3 years ago
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