Ask question

Ask question

All categories

4 years ago

10

If an object is moving with a constant velocity, are there any forces acting on the object?

1 answer:

Artist 52 [7]4 years ago

6 0

Answer:

there is no net force acting on the object

Explanation:

If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).

thank ya :>

You might be interested in

Internal Forces are :

Ugo [173]

Hi Pupil Here's your answer ::

➡➡➡➡➡➡➡➡➡➡➡➡➡

They are always balanced forces since if they are applied on an object the resultant of tgese forces on the object is Zero

So, the option A is the correct. They are always balanced forces.

8 0

3 years ago

A charged particle is surrounded by an electric field and a magnetic field

zlopas [31]

Answer:

yeah

Explanation:

electric fields help the charged particles interact

and isn't magnetic field the same as electric field.

7 0

3 years ago

Read 2 more answers

Pls help quickly

emmasim [6.3K]

Answer: frosted glass

Explanation:

8 0

3 years ago

Read 2 more answers

A planet has a period of revolution about the Sun equal to and a mean distance from the Sun equal to R .T^2 varies directly as _

shutvik [7]

T² caries directly as R³ .

This is Kepler's 3rd law of planetary motion .

4 0

3 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

3 years ago