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Harrizon [31]
1 year ago
12

A body is oscillating up and down at the end of a spring. Let’s consider when the body is at the top of its up-and-down motion.

Physics
1 answer:
Klio2033 [76]1 year ago
3 0

The velocity of the body is zero; option A

<h3>What is the motion of an oscillating body?</h3>

The motion of an oscillating body is known as simple harmonic motion.

Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.

For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.

In conclusion, oscillating bodies undergo simple harmonic motion.

Learn more about simple harmonic motion at: brainly.com/question/24646514

#SPJ1

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1.9 km converted to miles
Snowcat [4.5K]
When you convet km to miles this is what you get 

1.18061
7 0
3 years ago
What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

7 0
3 years ago
According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind
Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0
3 years ago
The density of silver is 10.5 g cm3. a piece of silver with a mass of 61.3 g would ovvupy a volume of ?
natali 33 [55]
The relationship between mass m, volume V and density d is:
d= \frac{m}{V}
The silver has density d=10.5 g/cm^3, and the mass of the piece of silver is m=61.3 g. Therefore we can calculate its volume using the previous formula:
V= \frac{m}{d}= \frac{61.3 g}{10.5 g/cm^3}=5.84 cm^3
7 0
2 years ago
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0
2 years ago
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