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Akimi4 [234]
3 years ago
8

What's the formula to calculate the radius of pendulum bob ​

Physics
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Answer:

The combination, L = I / (m * R) , that appears in the equation for the period of a physical pendulum, is called radius of oscillations

Hope this helps :]

You might be interested in
How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb kevlar canoe of the same size c
romanna [79]

We use the formula, to calculate the volume of water displaced by concrete canoe,

V =\frac{W}{\gamma }

Here, W is the weight of concrete canoe and \gamma is the specific weight of water and its value is 62.4\ lb/ft^3.

So,

V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3.

Now the volume of water occupied in ultra lightweight kevlar canoe,

v=\frac{w}{\gamma}

Here, w is weight of  kevlar canoe.

So,

v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3

Thus, the volume of water displaced,

=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3.

Hence, the volume of water displaced canoe compared to an ultra-lightweight  kevlar canoe is 2.19\ ft^3

5 0
3 years ago
An object is held 24.8 cm from a lens of focal length 16.0 cm. What is the magnification of the image?
Gwar [14]

Answer: 1.8

Explanation:

You are given

the object distance U = 24.8 cm

Focal length F = 16.0 cm

First find the image distance by using the formula:

1/f = 1/u + 1/v

Where V = image distance

Substitute u and f into the formula

1/16 = 1/24.8 + 1/v

1/ v = 1/16 - 1/24.8

1/v = 0.0625 - 0.04032258

1/v = 0.022177

Reciprocate both sides by dividing both sides by one

V = 45.09 cm

Magnification M is the ratio of image distance to the object distance. That is,

M = V/U

Substitute V and U into the formula

M = 45.09/24.8

M = 1.818

Magnification of the image is therefore equal to 1.8 approximately

3 0
3 years ago
The length of a wooden rod is 25.5 cm. What is this length in:<br>(a) millimetres?<br>(b) metres?​
larisa86 [58]

Answer:

(a)255mm \\ (b)0.255m

Explanation:

(a)1cm = 10mm \\ 25.5cm = x

Cross multiply

1x = 25.5 \times 10 \\ x = 255mm

(b)100cm = 1m \\ 25.5cm = x

cross multiply

100x = 25.5 \\  \frac{100x}{100}  =  \frac{25.5}{100}  \\ x = 0.255m

hope this helps

brainliest appreciated

good luck! have a nice day!

4 0
3 years ago
Read 2 more answers
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
2 years ago
Help!
Evgen [1.6K]

Answer:

factory:

-mechanical energy

-nuclear energy

-gravitational energy

Explanation:

8 0
3 years ago
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