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2 years ago

A simple pendulum of length 2.5 m makes 5.0 complete swings in 16 s. What is the acceleration of gravity at the location?

Physics

1 answer:

Norma-Jean [14]2 years ago

8 0

<h2>The acceleration of gravity at the location is 9.64 m/s²</h2>

Explanation:

Length of pendulum = 2.5 m

Time taken for 5 swings = 16 seconds

Time taken for 1 swing = 3.2 seconds

Period of pendulum = 3.2 seconds.

We have equation for period of simple pendulum as

$T=2\pi \sqrt{\frac{l}{g}}$

Where l is the length of pendulum and g is acceleration due to gravity.

Substituting

$T=2\pi \sqrt{\frac{l}{g}}\\\\3.2=2\pi \sqrt{\frac{2.5}{g}}\\\\g=\frac{4\pi^2 \times 2.5}{3.2^2}\\\\g=9.64m/s^2$

The acceleration of gravity at the location is 9.64 m/s²

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Consider two circular metal wire loops each carrying the same current I as shown below. In what r... Consider two circular metal

NeX [460]

Answer:

1) The magnetic field outside the loop is zero.

In region III the magnetic fields due to the two wire loops point in the opposite direction andhence cancel each other. Therefore the magnetic field is zero in region I, III and V

The diagram is attached

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2 years ago

Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and

masha68 [24]

Answer:

F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

fr = μ N

We define a reference system parallel to the floor

block B ( lower)

Y axis

N - W₁-W₂ = 0

N = W₂ + W₂

N = (M + m) g

X axis

F -fr = M a

for block A (upper)

X axis

fr = m a (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

F = (M + m) a (3)

a = $\frac{F}{ M+m}$

the friction force has the formula

fr = μ N

fr = μ (M + m) g

let's calculate

fr = 0.34 (2.0 + 0.250) 9.8

fr = 7.7 N

we substitute in equation 2

fr = m a

a = fr / m

a = 7.7 / 0.250

a = 30.8 m / s²

we substitute in equation 3

F = (2.0 + 0.250) 30.8

F = 69.3 N

5 0

2 years ago

Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

Nadya [2.5K]

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

$PE=(8)(1.6)(2)=25.6 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

2 years ago

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At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init

Lady_Fox [76]

Answer:

$T_{f}$ = 85.89 ° C

Explanation:

The linear thermal expansion process is given by

ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

ΔV = V β ΔT

Let's apply this equation to our case

ΔV / V = -0.507% = -0.507 10-2

ΔT = (ΔV / V) 1 /β

ΔT = -0.507 10⁻² 1 / 1.15 10⁻³

ΔT = -4.409

$T_{f}$ –T₀ = 4,409

$T_{f}$ = T₀ - 4,409

$T_{f}$ = 90.3-4409

$T_{f}$ = 85.89 ° C

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2 years ago

Read 2 more answers