All categories

3 years ago

What is the approximate diameter of an inflated

1 answer:

6 0

Well let's convert all these values out of standard form first:

2x10^-2 = 2x0.01 = 0.02m = 2cm
2x10^0m = 2x1 = 2m
2x10^-1m = 2x0.1 = 0.2m = 20cm
2x10^1m = 2x10 = 20m

Based on that, we know that 20cm is roughly equivalent to a basketball (at least it's closer than all the other values), so the answer is therefore 2 - 2x10^-1m

You might be interested in

An unwary football player collides with a padded goalpost while running at a velocity of 9.50 m/s and comes to a full stop after

Amanda [17]

Answer:

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

Explanation:

Given:

Initial velocity of the player (u) = 9.50 m/s

Final velocity of the player (v) = 0 m/s (Comes to a stop)

Displacement of the player (S) = 0.250 m

We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:

$v^2=u^2+2aS$

Expressing in terms of 'a', we get:

$a=\frac{v^2-u^2}{2S}$

Plug in the given values and solve for 'a'. This gives,

$a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2$

Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².

Now, using the first equation of motion, we have:

$v=u+at\\\\t=\frac{v-u}{a}$

Plug in the given values and solve for 't'. This gives,

$t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s$

Therefore, the the collision will last for 0.053 s.

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

8 0

3 years ago

A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inel

kkurt [141]

Answer:

7.5 m/s

Explanation:

Unfortunately, I don't have an explanation but I guessed the correct answer.

4 0

3 years ago

Help me please I only have a few minutes to get this doneplease and thank q

DIA [1.3K]

A: particles are more spread out in gas

6 0

3 years ago

Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth.

bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

$f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}$

3 0

3 years ago

HELP ASAP!! WILL MARK BRAINLIEST

Kitty [74]

Answer:

Explanation:

The magnet can always attract other things.

8 0

3 years ago

Read 2 more answers