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3 years ago

Acceleration toward the center of a curved path is called

Physics

1 answer:

Serggg [28]3 years ago

7 0

Answer:

Centripetal acceleration.

Explanation:

Centripetal acceleration is a property of a body moving in a uniform circular path and it is directed radially towards the center of the circle in which body is rotating.

The force which causes this acceleration is centripetal force which is also directed towards the center of the circle and pulls the body towards its center.

It is calculated through following formula

$a=v^2/r$

where v is velocity and r is the radius of the circle.

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An atom with the expected number of neutrons, protons, and electrons is called a(n)

ryzh [129]

Answer:

Stable atom

Explanation:

A stable atom is one that has a balanced nuclear inter-particle force reaction as such the binding energy of a stable atom is sufficient to permanently keep the nucleus as one unit. Examples of a stable atom are the atoms of monoisotopic elements such as fluorine, sodium, iodine, gold, aluminium, and cobalt.

In a stable atom the expected number of proton, neutron, and electron are present while in an unstable atom or radioactive atom, there are more than the expected number of neutrons or protons, such that the internal energy of the nucleus is excessive and more than the binding energy, which can lead to radioactive decay.

6 0

3 years ago

A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov

Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

3 0

2 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

Question 7(Multiple Choice Worth 4 points)

Nimfa-mama [501]

C is the correct answer, as the equation for photosynthesis is carbon dioxide +water=glucose(sugar) + oxygen

3 0

3 years ago

Read 2 more answers

Q11) If you were standing at the top of a building and you dropped a rock.

Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0

2 years ago