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Over [174]
3 years ago
13

"a uniform thin rod of length 0.962 m is hung from a horizontal nail passing through a small hole in the rod located 0.048 m fro

m the rod's end. when the rod is set swinging about the nail at small amplitude, what is the period of oscillation?"
Physics
1 answer:
exis [7]3 years ago
5 0
<span>1.57 seconds. The rod hanging from the nail constructs a physical pendulum. The period of such a pendulum follows the formula T = 2*pi*sqrt(L/g) where T = time L = length of pendulum g = local gravitational acceleration So the problem becomes one of determining L. It's tempting to consider L to be the distance between the center of mass and the pivot, but that isn't the right value. The correct value is the distance between the pivot and the center of percussion. So let's determine what that is. We can treat the uniform thin rod as an uniform beam and for an uniform beam the distance between the center of mass and the center of percussion is expressed as b = L^2/(12A) where b = distance between center of mass and center of percussion L = length of beam A = distance between pivot and center of mass Since the rod is uniform, the CoM will be midway from either end, or 0.962 m / 2 = 0.481 m from the end. The pivot will therefore be 0.481 m - 0.048 m = 0.433 m from the CoM Now let's calculate the distance the CoP will be from the CoM: b = L^2/(12A) b = (0.962 m)^2/(12 * 0.433 m) b = (0.925444 m^2)/(5.196 m) b = 0.178107005 m With the distance between the CoM and CoP known, we can now calculate the effective length of the pendulum. So: 0.433 m + 0.178107005 m = 0.611107005 m And finally, with the effective length known, let's calculate the period. T = 2*pi*sqrt(L/g) T = 2*pi*sqrt((0.611107005 m)/(9.8 m/s^2)) T = 2*pi*sqrt(0.062357858 s^2) T = 2*pi*0.249715554 s T = 1.569009097 s Rounding to 3 significant figures gives 1.57 seconds. Let's check if this result is sane. Looking up "Seconds Pendulum", I get a length of 0.994 meters which is longer than the length of 0.611 meters calculated. But upon looking closer at the "Seconds Pendulum", you'll realize that it's period is actually 2 seconds, or 1 second per swing. So the length of the calculated pendulum is sane.</span>
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Insulators have very high .
Vlad1618 [11]

Answer:

Resistance to electrical currents

Explanation:

Conductors have low resistance to electrical currents, and are used to "conduct" the flow of electricity.

Insulators have very high resistance and are used to protect us from the flow of electricity.

5 0
3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
A charge q produces an electric field of strength 2E at a distance of d away. Determine the electric field strength at a distanc
Alja [10]

The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.

If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²).  That's (2E)/4 = 0.5E .

3 0
3 years ago
Read 2 more answers
BRAINLEST FOR CORRECT ANSWER
xxMikexx [17]

Answer:

3kg sledgehammer swung at 1.5 m/s

Explanation:

Small Sledgehammer:

Mass:3.0

Velocity:1.5

MASS×VELOCITY=MOMENTUM

3.0×1.5= 4.5 (momentum)

Large Sledgehammer:

Mass:4.0

Velocity:0.9

4.0×0.9=3.6 (momentum)

higher momentum is the smaller Sledgehammer.

3 0
3 years ago
A force of 8,480 N is applied to a cart to decelerate it at a rate of 32.0 m/s 2 . What is the mass of the cart?
Vinvika [58]
The force is given by the product of mass and the acceleration thus the force is given
8480/32 is equal to 265 kg
5 0
3 years ago
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