explain why the motion of a body which is moving with constant speed in a circular path is said to be accelerated ?

Answers

Answer 1

Answer: In this case when a body moves in circular path acceleration is called Centripetal acceleration... Body in circular path moves with constant speed but its direction changes at every instant of time ...as acceleration is produced when either magnitude or direction of velocity changes...here magnitude remains same but direction changes so body is said to be accelerated....

Answer 2

Answer: as its velocity changes and hence is known as accelarated

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Can any kind soul help me please

Answers

Explanation:

Hey there!

Given;

Mass of the Earth (M) = 6.0*10^24kg.

Radius of Earth (R) = 6400km = 6400*1000= 6.4*10^6.

Gravitational constant (G) = 6.67*10^-11Nm^2/kg^2.

Height from surface of Earth (h) = 350km= 3.5*10^5m

Gravity (g) = ?

We have;

$g = \frac{G.M}{ {(R + h)}^(2) }$

Where "G" is gravitational constant, "M" is mass of earth, "g" is acceleration due to gravity, "h" height from surface of Earth.

Keep all values.

$g = \frac{6.67 * {10}^( - 11) * 6.0 * {10}^(24) }{ (6.4 * {10}^(6) + 3.5 * {10}^(5) )^2}$

$or \: g = \frac{40.002* {10}^(13) }{ (6.4* {10}^(6) + 3.5*10^5)^2}$

g=8.7m/s^2

Therefore,gravityfromdistance350kmabove theEarth'ssurfaceis8.7m/s^2.

Hopeithelps...

HELP NEEDED! Determine the mechanical energy of this object: a 1-kg ball rolls on the ground at 2 m/s. A. The ball is on the ground, so the mechanical energy is zero.
B. 2 J
C. 1 J
D. 4 J

Answers

The mechanical energy of an object is the sum of its potential and kinetic energy.

Because the ball is on the ground, its potential energy is 0.
Its kinetic energy is given by:

K.E = 1/2 mv²
K.E = 1/2 x 1 x 2²
K.E = 2 J

Mechanical energy = 2 + 0
Mechanical energy = 2 J

The answer is B.

I think it's 2 J. Hope I helped :)

The wavelength of a wave doubles as it travelsfrom medium A into medium B. Compared to the
wave in medium A, the wave in medium B has
(1) half the speed
(2) twice the speed
(3) half the frequency
(4) twice the frequency

Answers

Final answer:

The situation described in the question pertains to wave propagation. Given that wave speed in a given medium is constant, and the relationship between wave speed, wavelength, and frequency, the wave in medium B, which has double the wavelength of the wave in medium A, would have half the frequency of the wave in medium A.

Explanation:

This question is about wave propagation between two different mediums, labeled 'A' and 'B'. The fundamental principle involved here is the dependence of wave speed on the medium through which it travels. However, it's necessary to recognize that wave speed also depends on wavelength and frequency. This relationship is encapsulated by the formula v = fλ, where v is wave speed, f is frequency, and λ is wavelength.

From the problem statement, we understand that the wavelength doubles when going from medium A to medium B. We are then asked to deduce the corresponding change in wave speed or frequency. If the wavelength doubles and the wave speed remains the same, the only possible adjustment is for the frequency to halve. So, the wave in medium B, compared to the wave in medium A, would have half the frequency (3). This is because wave speed in a given medium is constant and as per the formula, if wavelength increases, frequency must decrease.

Learn more about Wave Propagation here:

brainly.com/question/31084304

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Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is traveling in the same direction at 25 km/hr. Which of the following statements is true?A) The momentum before the collision is equal to the momentum after the collision.

B) The momentum before the collision is greater than the momentum after the collision.

C) The momentum before the collision is less than the momentum after the collision.

Answers

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_(1) \ and\ m_(2)$

Let the velocity of cars A and B are denoted as $v_(1) \ and\ v_(2)$

The momentum before collision is-

$p_(i) =m_(1) v_(1) +m_(2) v_(2)$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_(1) \ and\ v'_(2)$

Hence the final momentum of the system is-

$p_(f) = m_(1) v'_(1) +m_(2) v'_(2)$

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e

$p_(i) =p_(f)$

$m_(1) v_(1) +m_(2)v_(2) =m_(1) v'_(1) +m_(2) v'_(2)$

Hence the option A is right.

A
This principle is known as the law of conservation of momentum.

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.a. At what speed was the stone launched?
b. What is the speed and angle of impact?

Answers

As per the question a stone is projected horizontally from the top of a 20 meter cliff.

Hence the height [h] of the cliff = 20 m.

The stone lands 36.0 m away.

Hence horizontal distance R] = 36 m.

First we are asked to calculate the launching speed of the stone.

Let it be denoted as u.

The motion of particle is the force of gravity only.Hence it is a projectile motion.

The total time taken by a projectile when it is fired horizontally at a height h from the ground is -

$T=\sqrt{(2h)/(g) }$ [ where g is the acceleration due to gravity = 9.8 m/s^2 or 10 m/s^2]

$=\sqrt{(2*20)/(10) } s$

$= 2 s$

The range of the projectile will be-

$R= \ horizontal\ speed*\ total\ time\ of\ flight$

The horizontal speed for a projectile is uniform through out the motion of the projectile as long as gravity is constant.

In this situation the horizontal velocity is equal to the initial projected speed i.e u.

Hence horizontal distance R = u ×T

Putting the value or R and T in the above formula we get-

$R=u*\sqrt{(2h)/(g) }$

$36 = u*\sqrt{(2*20)/(10) }$

$36 = u*2$

$u =(36)/(2)$

$u =18 m/s$

Now we are asked to calculate the angle of impact.

Initially the vertical velocity of the particle is 0. The path of the trajectory of the particle will be parabolic due to the force of gravity.At any instant of time the horizontal component of the instantaneous velocity of the of the particle will be constant .

Let the horizontal component is denoted as $V_(x) \ where \ V_(x) = u$

The vertical component is calculated as follows-

We know that v = u +at

Here v is the final velocity, u is the initial speed, a is the acceleration and t is the instantaneous time.

For vertically downward motion under gravity u = 0 and a = -g

Let the vertical velocity at any instant is denoted as $V_(y)$

Hence

$V_(y) = 0 -gt$

$V_(y) = gt$ [Here we are taking only the magnitude]

Let the resultant velocity makes an angle α with the horizontal when it strikes the ground.

The horizontal and vertical component is denoted in the diagram below-

From the figure we see that -

$tan\alpha =(V_(y) )/(V_(x) )$

$tan\alpha =(gt)/(u)$

$tan\alpha =(10*2)/(18)$

$tan\alpha =1.11111$

$\alpha =48.01^0$ [ans]

Hence the angle of impact is close to 48 .1 degree .

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_(o)+v_(o)t+ (at^2)/(2) \n 20= (10t^2)/(2) \n t=2s$

Through Definition of Velocity, comes:

$\Delta v= (\Delta S)/(\Delta t) \n v_x= (36)/(2) \n \boxed {v_(x)=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_(y)=v_{y_(o)}+gt \n v_(y)=10*2 \n \boxed {v_(y)=20m/s}$

The angle of impact is given by:

$cos(\theta) =(v_(x))/(v_(y)) \n cos(\theta) = (18)/(20) \n cos(\theta) =0.9 \n arccos(0.9)=\theta \n \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the index of refraction? A)0.827
B)0.944
C)1.21
D)1.06

Answers

The equation for the index of refraction is n = sin(theta1)/sin(theta2) where theta1 is the refracted angle and theta2 is the angle in a medium. Plugging in the values for the index of refraction, n = sin(32)/sin(26) we get 1.21. the answer is letter C. The index of refraction is always greater than 1.

That answer is wrong it is not 1.21 just turned it in and was wrong

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