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3 years ago

Why does an object traveling in a circle at a constant speed always accelerate ?

Physics

2 answers:

Taya2010 [7]3 years ago

8 0

I'm not sure but I believe it is momentum, when an object keeps moving and keeps Its speed

LUCKY_DIMON [66]3 years ago

6 0

Because "acceleration" doesn't mean "speeding up". It means any change
in the speed or direction of motion.

An airplane speeding up, a bicycle slowing down, and a car going around a curve, are all doing accelerated motion.

A circle has no straight parts, so an object traveling in a circle is always changing
its direction. That means accelerated motion, even if its speed doesn't change..

You might be interested in

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

maria [59]

Answer:

a) a = 5.03x10¹³ m/s²

b) $V_{f} = 4.4 \cdot 10^{5} m/s$

Explanation:

a) The acceleration of the positron can be found as follows:

$F = q*E$ (1)

Also,

$F = ma$ (2)

By entering equation (1) into (2), we have:

$a = \frac{F}{m} = \frac{qE}{m}$

Where:

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C

E: is the electric field = 286 N/C

$a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2}$

b) The positron's speed can be calculated using the following equation:

$V_{f} = V_{0} + at$

Where:

$V_{f}$ : is the final speed =?

$V_{0}$ : is the initial speed =0

t: is the time = 8.70x10⁻⁹ s

$V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s$

I hope it helps you!

4 0

3 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

During which type of collision is none of the energy converted to sound

rewona [7]

I believe the answer is C

4 0

3 years ago

When part of the rock record is destroyed, the erosional gap that forms is

zaharov [31]

Answer:

D superposition

Explanation:

sana makatulong:)

8 0

3 years ago

Two objects, each with a charge of +0.15 C are separated by a distance of 3 meters.

Mrac [35]

Answer:

A . Are the masses repelling or attracting

answer:

So, positive energy densities ("masses") attract each other by gravitational interaction. This is the general idea.

B. What is the magnitude of the electrical force between the objects?

answer:

 The magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

C. What would the magnitude of the force be if one of the charges was 1/3 the amount?

answer:

0.45

D. ) What would the force be if the distance between the charges was only 1 meter (with the original charges)?

answer;

Fmin=2.3×10^−28N

E: What would the force be if one of the charges was 1/3 the amount AND the distance was 1 meter?

answer;

Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.

#CARRYONLEARNING:)

LOVEUALL

6 0

3 years ago