a 90 gram object moves around a circle whose diameter is 2.56 m at a constant speed off 16 m/s. what is the object acceleration? how much time will take to complete the around?

Answers

Answer 1

Answer:

[tex]a=200\ m/s^2[/tex] and t = 0.5 s

Explanation:

Given that,

Mass of the object, m = 90 gram = 0.09 kg

Diameter of the circle, d = 2.56 m

Radius, r = 1.28 m

Speed, v = 16 m/s

We need to find the object acceleration. On the circular path, acceleration is given by :

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(16)^2}{1.28}\\\\a=200\ m/s^2[/tex]

So, the object's acceleration is [tex]200\ m/s^2[/tex].

Let t is time it will take to complete the around. Speed is given by :

[tex]v=\dfrac{2\pi r}{t}\\\\t=\dfrac{2\pi r}{v}\\\\t=\dfrac{2\pi \times 1.28}{16}\\\\t=0.5\ s[/tex]

So, time to cover one complete round is 0.5 s.


Related Questions

1. An object carries a charge of q = +4x10-8C.
How many electrons are needed to make it elec-
trically neutral?​

Answers

Answer: i can help you

Explanation:

[tex]2.5[/tex]×[tex]10^{11}[/tex] electrons are needed to make an object neutral which is carrying a charge of [tex]q=+4[/tex]×[tex]10^{-8}[/tex]

What is a charge of an electron?The elementary charge (e) has a negative sign, and the electron's charge is equal to its magnitude. The charge of the electron is -1.602 x [tex]10^{-19}[/tex] coulombs (C), while the elementary charge is approximately 1.602 x [tex]10^{-19}[/tex] coulombs (C).Thus, -e can be used to represent the charge of an electron. The proton only possesses an e-charge despite being far more massive than the electron. The number of protons and electrons in neutral atoms is constant.The electron was unquestionably discovered by JJ Thomson. He only succeeded in obtaining the electron's charge-to-mass ratio, though, despite all the experiments he conducted on it. Robert Millikan holds the distinction of being the first to quantify the charge on an electron through his oil-drop experiment in 1909.

To learn more about charge, refer to

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What can be described as a technology that uses high-frequency sound waves to create images?

A. ultrasound
B. echolocation
C. analog signal
D. digital signal

Answers

Answer:

Ultrasound can be described as a technology that uses high-frequency sound waves to create images.

Explanation:

Ultrasound can be described as a technology that uses high-frequency sound waves to create images.

What is ultrasound?

Ultrasound imaging uses sound waves to produce pictures of the inside of the body. It helps diagnose the causes of pain, swelling and infection in the body's internal organs and to examine an unborn child (fetus) in pregnant women.

In infants, doctors commonly use ultrasound to evaluate the brain, hips, and spine. . It also helps guide biopsies, diagnose heart conditions, and assess damage after a heart attack. Ultrasound is safe, non-invasive, and does not use radiation.

This procedure requires little to no special preparation. Your doctor will tell you how to prepare, including whether you should not eat or drink beforehand. Leave jewellery at home and wear loose, comfortable clothing. You may need to change into a gown.

To know more about ultrasound follow

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hi people see bye 18 45 36

Answers

Answer:

bye

Explanation:

Which statement best describe one of George Berkeley's arguments against materialism?
O A. There cannot be two different substances; my mind exists; therefore, matter does not exist.
OB. To be is to be perceived; we don't perceive matter; therefore, matter does not exist.
OC. I can only perceive my own mind; therefore, matter does not exist.
OD.
I can only perceive other minds; therefore, matter does not exist.
Reset
Next

Answers

Answer:

the answer is B

Explanation:

materialism is just wanting to always buy new things B does the opposite

A car is traveling at 70 mi/hr. The car travels for half an hour, how far has the car gone? Don't forget to include units of measurement.

Answers

Distance = (speed) x (time)

Distance = (70 mi/hr) x (0.5 hr)

Distance = 35 miles

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

Answers

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  [tex]\tau = 34.3 \ N\cdot m[/tex]

Explanation:

From the question we are told that

   The mass of the steel ball is  [tex]m = 3.0 \ kg[/tex]

    The length of arm is  [tex]l = 70 \ cm = 0.7 \ m[/tex]

    The mass of the arm is [tex]m_a = 4.0 \ kg[/tex]

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       [tex]r = \frac{l}{2}[/tex]

=>    [tex]r = \frac{ 0.7}{2}[/tex]  

=>    [tex]r = 0.35 \ m[/tex]  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      [tex]\tau = m_a * g * r + m * g * L[/tex]

=>    [tex]\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70[/tex]

=>    [tex]\tau = 34.3 \ N\cdot m[/tex]

What is the ratio of a 0.7 in circle in centimeters.

Answers

I believe it would be 4.4

A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bicyclist’s acceleration to the nearest tenth of a m/s^2 ?

Answers

Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

[tex]u = 2.5m/s[/tex] ---- Initial Velocity

[tex]v = 12.5m/s[/tex] ---- Final Velocity

[tex]t = 4.5s[/tex] ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

[tex]v = u + at[/tex]

Substitute values for v, u and a

[tex]12.5 = 2.5 + a * 4.5[/tex]

[tex]12.5 = 2.5 + 4.5a[/tex]

Collect Like Terms

[tex]4.5a = 12.5 - 2.5[/tex]

[tex]4.5a = 10.0[/tex]

Solve for a

[tex]a = 10.0/4.5[/tex]

[tex]a = 2.2m/s^2[/tex] ---- (approximated)

Hence, the bicyclist's acceleration is 2.2m/s^2

Some help me on this please


What can be said for certain about the affect of an external net force on
an object? *

A.It changes the object's mass.

B.It changes the object's kinetic energy.

C.It changes the object's temperature.

D.It changes the object's momentum.

Answers

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

if air had a taste, would we all not need to eat again?

Answers

Answer:

prolly

Explanation:

In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary second ball of mass 0.4 kg. What is the velocity of the 0.3 kg ball after the collision? Assume collision is elastic.

Answers

Answer:

The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Explanation:

Given;

mass of the first object, m₁ = 0.3 kg

initial velocity of the first ball, u₁ = 2.8 m/s

mass of the second ball, m₂ = 0.4 kg

initial velocity of the second ball, u₂ = 0

let the final velocity of the first ball, = v₁

let the final velocity of the second ball, = v₂

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂

0.84 = 0.3v₁ + 0.4v₂

2.8 = v₁ + 1.333v₂ -------equation (1)

Apply one-direction velocity;

u₁ + v₁ = u₂ + v₂

2.8 + v₁ = 0 + v₂

v₂ = 2.8 + v₁

substitute the value of v₂ into equation (1)

2.8 = v₁ + 1.333v₂

2.8 = v₁ + 1.333(2.8 + v₁)

2.8 = v₁ + 3.732 + 1.333v₁

2.8 - 3.732 = v₁ + 1.333v₁

-0.932 = 2.333v₁

v₁ = -0.932 / 2.333

v₁ = -0.4 m/s

Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

1.) True or False. Scientists have drilled through the crust to reach the mantle.

Answers

Answer:

False

Explanation:

We have never made it to the mantle.

False. That is the correct answer
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