Two friends are walking on a hot beach on a hot summer day. one friend reports her feet are becoming hot she needs her sandals.

What is the most appropriate SI unit to express the speed of a cyclist in a 10-km race? km/s cm/h km/h mm/s

Murljashka [212]

You sure wouldn't want something like cm/s or (yikes cm/hr). You want a reasonable number for sports usually between 0 and 100

Km / hour would be a good choice.

The next town to where I live is 25 km away. On a good day, I can make it there in about 3/4 of an hour.

Speed = 25 km / 0.75 hour = 33.3 km/hour. That's actually a little fast most of the time. But you should understand what I mean.

5 0

3 years ago

Why does a vibrating simple pendulum not to produce any sound

Gelneren [198K]

It does produce 'sound' ... a compression wave traveling through the air. But your ears don't hear a sound that's vibrating less than 20 or 30 times every second. If you could swing your pendulum that fast, you could hear the sound of its vibrations pushing the air around.

7 0

3 years ago

The half-life of iodine-131 is 8 days. Chan Hee examines two samples of iodine-131. Sample 1: 10 g Sample 2: 20 g What can Chan

Illusion [34]

Chan Hee can expect to observe about the samples after 16 days is that three-fourths of each sample will have undergone radioactive decay. The best answer is letter B.

4 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

A PERSON GETTING OUT OF MOVING BUS FALLS IN THE DIRECTION OF MOTION OF THE BUS. WHY?

djverab [1.8K]

Answer:

PLEASE MARK AS BRAINLIEST!!

Explanation:

A getting passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion.

3 0

2 years ago

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