Ask question

Ask question

All categories

2 years ago

11

Two or more velocities add by

1 answer:

forsale [732]2 years ago

6 0

Two or more velocities add by vector sum.
In fact, velocity is a vector, with its own intensity, direction and verse. Therefore, we cannot simply add two velocities by algebraic sum, but we need to decompose them on two axes (if we are in a 2D-plane) and calculate the resultant on each axis, and then calculate the total resultant of the two vectors, which will have its intensity, direction and verse.

You might be interested in

Describe a change caused by kinetic energy as well as a change that involves potential energy

kykrilka [37]

Kinetic energy is formed when the object is in motion.
Potential energy is the energy that is formed relative to others.

One of the example is Corn flour factory.

Corn turned into flour by a windmill that moved by the waterfall. Movement of the mill is relative to the power given by waterfall (potential energy) and the spinning crushes the corn into flour (kinetic energy)

5 0

2 years ago

Read 2 more answers

A car is traveling at 70 km/h. It then uniformly decelerates to a complete stop in 12 s. Find its acceleration ( in m/s2 ).

Yakvenalex [24]

Answer:

Acceleration will be $-1.620m/sec^2$

Explanation:

We have given initial speed of the car is 70 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec$

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So $0=19.444+a\times 12$

$a=\frac{-19.444}{12}=-1.620m/sec^2$ ( negative sign indicates that speed of the car will constantly decrease )

5 0

3 years ago

Read 2 more answers

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

5 0

3 years ago

A small button placed on a horizontal rotating platform with diameter 0.320 m will revolve with the platform when it is brought

Otrada [13]

Answer:

0.2687 approximately 0.27

Explanation:

Diameter = 0.320

Speed = 40.0 rev/min

We are required to find coefficient of static friction between friction and button

The radius can be calculated as

0.320/2

= 0.160m

Then we have the rotational speed w = 40rev/min x 2pi/60

= 4.19 rad/s

umg = mrw²

u = mrw²/mg

u = rw²/g -------(1)

g = 9.8

When we put values into equation 1

0.150m x 4.19² / 9.8

= 0.150m x 17.5561 /9.8

= 0.2689

This is approximately 0.27

6 0

2 years ago

A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

SSSSS [86.1K]

I had the same question

6 0

2 years ago