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zimovet [89]
1 year ago
7

1. thomas jefferson proposed using the length (l) of a simple pendulum whose period (t) was exactly 2 seconds as the definition

of a standard meter. what is the length (l) of a 2 sec. pendulum in si units?
Physics
1 answer:
Alecsey [184]1 year ago
7 0

The length of a 2 sec pendulum is 1 m.

Given that, initial length of the simple pendulum L₁ = 1 m

Initial time period T₁ = 2 sec

We need to find the length of the pendulum whose time period is 2 sec

T₂ = 2 sec

L₂ = ?

We know that the time period of the simple pendulum is given by the formula,

T = 2π√(L/g)

From the above relation, we can write T ∝ √L

T₁ / T₂ = √(L₁/L₂)

Making L₂ from the above relation, we have,

L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m

Thus, the length of a 2 sec pendulum is 1 m.

To know more about time period:

brainly.com/question/17350379

#SPJ4

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Ivanshal [37]

Answer:

The Moon's distance from the Earth varies during its orbit

Explanation:

The correct statement is ,The Moon's distance from the Earth varies during its orbit.

Important point regarding moon:

1 .Moon is a natural satellite of the earth.

2. Moon is the fifth largest satellite in solar system.

3.Second densest  satellite in solar system.

4.Moon rotates about earth.

5.Moon is an astronomical body .

5 0
3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires
svlad2 [7]

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

Y=\frac{F \times L}{A\times \delta L}

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

8 0
2 years ago
A.to reduce<br> b.to dispose<br> c.to prevent<br> d.to help
kenny6666 [7]

Answer:

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5 0
3 years ago
What is the motion of the object?
aleksley [76]

Answer:

<em>Thus, the object is accelerating to the left</em>

Explanation:

<u>The Net Force</u>

The net force is the result of adding all the forces as vectors acting on a body.

\vec F=\vec F_1+\vec F_2+...+\vec F_n

Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.

Second Newton's law gives the relation between the net force and the acceleration of the body:

\vec F = m.\vec a

We can see the acceleration is a vector with the same direction as the net force.

The diagram shows two vertical forces and two horizontal forces.

The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.

The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N

Thus, the object is accelerating to the left

4 0
3 years ago
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

7 0
3 years ago
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