All categories

3 years ago

Going around a ferris wheel without stopping is an example of _____ circular motion(apex)

Physics

2 answers:

Natasha_Volkova [10]3 years ago

5 0

Explanation:

Going around a Ferris wheel without stopping is an example of <u>uniform</u> circular motion. In a circular motion, the objects moves with constant speed but its velocity is continuously changing because of changing direction.

The force acting when the object when it is moving in a circular motion is called centripetal force. The direction of both force and acceleration is inwards.

Hence, "going around a Ferris wheel without stopping" is an example of uniform circular motion.

victus00 [196]3 years ago

3 0

going around a ferris wheel without stopping is an example of uniform circular motion (apex).

You might be interested in

If you set up set up your account to use a security key, can you still get in without having the physical key ?

butalik [34]

Answer:

yes \\

have a nice day T_T

8 0

2 years ago

What are three physical properties of gases

NikAS [45]

Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.

Explanation:

7 0

2 years ago

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

3 years ago

in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo

guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

$P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s$

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

4 0

3 years ago

A football player runs directly down the field for 45m before turning to the right at an angle of 25 degrees from his original d

Fiesta28 [93]

............................................................................... Hello wonderful person <3

5 0

2 years ago