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4 years ago

Which example identifies a change in motion that produces acceleration?

2 answers:

3 0

The answer is the option a. <span>a. a ball moving at a constant speed around a circular track.

Acceleration is the change in velocity. This change may be either on speed or on direction (or both).

A ball moving at a constant speed around a circular track is continously changing its direction so it is under acceleration (centripetal acceleration).

All the other cases are of objects moving at the constant speed and in straight line, i.e. constant velocity, which is not accelerating.
</span>

Mekhanik [1.2K]4 years ago

3 0

Option (a) is correct. The motion of a ball moving in a circular path at a constant speed is termed as the motion with acceleration.

Explanation:

The acceleration of a body is defined as the rate of change of velocity of the body. If the velocity of object under motion continues to change during the motion, the object is considered to be moving under acceleration.

The change in velocity is not only about the change in the magnitude of the speed of the object but it is also the change in the direction of motion of the object moving at a constant speed.

The motion of a particle in a circular path at a steady speed is an accelerated motion because the direction of motion of the ball changes at every instant during its motion in a circular path. The change in direction also changes the velocity of the object and therefore, it is categorized as the accelerated motion.

The motion of a particle at constant velocity in vacuum, a skater moving with a constant speed on a straight track and a vehicle moving on a street at steady speed are not considered as the accelerated motion because the velocity of the body does not change and the direction of motion also remains the same during the motion.

Thus, Option (a) is correct. The motion of a ball moving in a circular path at a constant speed is termed as the motion with acceleration.

Learn More:

1. The acceleration of the bear brainly.com/question/6268248

2. The kinetic energy of an object depends on brainly.com/question/137098

3. Type of mirror used by dentist brainly.com/question/997618

Answer Details:

Grade: High School

Subject: Physics

Chapter: Motion in one dimension

Keywords:

Acceleration, velocity, rate of change, speed, direction of motion, steady speed, straight track, circular track, constant, vacuum.

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A) A 5.75 mL sample of mercury has a measured mass of 77.05 g. The density is ___________.

Kitty [74]

Answer:

Density of the sample will be 13.4 kg/L

Explanation:

We have given volume of the sample $V=5.75mL=5.75\times 10^{-3}L$

Mass of the sample $M=77.05gram =77.05\times 10^{-3}kg$

We have to find the density of the sample

Density of the sample is given by

$Density=\frac{mass}{volume}=\frac{77.05\times 10^{-3}}{5.75\times 10^{-3}}=13.4kg/L$

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4 years ago

A satellite has a mass of 5850 kg and is in a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The perio

koban [17]

Answer:

W = 24.28 kN

Explanation:

given,

Mass of satellite = 5850 Kg

height , h = 4.1 x 10⁵ m

Radius of planet = 4.15 x 10⁶ m

Time period = 2 h

= 2 x 3600 = 7200 s

Time period of satellite

$T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}$

R is the radius of planet

h is the height of satellite

$T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}$

now calculation of acceleration due to gravity

$g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}$

$g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}$

g = 4.15 m/s²

True weight of satellite

W = m g

W = 5850 x 4.15

W = 24277.5 N

W = 24.28 kN

True weight of the satellite is W = 24.28 kN

8 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

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