All categories

2 years ago

According to recent research, ice skaters are able to glide smoothly across the ice because _____.

Physics

1 answer:

nekit [7.7K]2 years ago

8 0

In the blank should go of friction.

You might be interested in

A block lies on a horizontal frictionless surface and

zhenek [66]

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

$Fx=\dfrac{1}{2}kx^2$

$\\\Rightarrow x=\dfrac{2F}{k}$

$\\\Rightarrow x=\dfrac{2\times 3}{60}$

$\\\Rightarrow x=0.1\ m$

The position of the block is 0.1 from the initial position.

6 0

2 years ago

Which of the following is the best description of a gaseous substance?

sveta [45]

Answer:

It has no shape of its own but has a definite volume.

Explanation:

Gases have no shape but a definite volume

6 0

2 years ago

Read 2 more answers

How far away is mars?

Elanso [62]

Answer:

about: 110.14 million mi

Explanation:

the distance to Mars from Earth is 140 million miles (225 million kilometers).But, distance to Mars from Earth is constantly changing.

Hope that was helpful.Thank you!!!

3 0

3 years ago

Read 2 more answers

This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the

valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2$

Equation of motion

$v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s$

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

5 0

3 years ago

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What

kati45 [8]

Explanation:

The given data is as follows.

radius (r) = 3.25 cm, $\alpha = 11.6 rad/s^{2}$

Now, we will calculate the tangential acceleration as follows.

$a_{tangential} = \alpha \times r$

Putting the given values into the above formula as follows.

$a_{tangential} = \alpha \times r$

= $11.6 rad/s^{2} \times 3.25 cm$

= 37.7 $rad cm/s^{2}$

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 $rad cm/s^{2}$ .

6 0

3 years ago