What is the freezing point of the substance in the test tube?

A ball of 0.1kg is thrown straight up into the air with

Contact [7]

a) The momentum of the ball at the maximum height is zero

b) The momentum of the ball at halfway is 1.06 kg m/s

Explanation:

a)

There is only one force acting on the ball during its motion: the force of gravity, acting downward. As a result, the ball hasa constant acceleration downward ( $g=9.8 m/s^2$ , acceleration of gravity), and therefore it is in free fall motion.

This means that the velocity of the ball is given by the equation:

$v=u-gt$

where

v is the velocity at time t

u = 15 m/s is the initial velocity

$g=9.8 m/s^2$

As we see from the equation, the term $-gt$ is negative: this means that as the ball goes higher and higher, its velocity decreases. The point of maximum height is reached when the velocity has become zero (after that point, the velocity changes direction and points downward), so when

$v=0$

The momentum of the ball at any point is given by

$p=mv$

where m = 0.1 kg is its mass. Therefore, at the maximum height, since $v=0$ , the momentum is also zero:

$p=0$

b)

We can use the following suvat equation to find the maximum height reached by the ball:

$v^2-u^2=2as$

where

v = 0 is the velocity at the point of maximum height

u = 15 m/s is the initial velocity

$a=-g=-9.8 m/s^2$ is the acceleration (downward)

s is the vertical displacement (the maximum height)

Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(15)^2}{2(-9.8)}=11.48 m$

This means that the ball is halfway when its height is

$h'=\frac{s}{2}=\frac{11.48}{2}=5.74 m$

We can find the velocity of the ball v' when it is at h' by using again the same equation:

$v'^2-u^2=2ah'\\v=\sqrt{u^2+2ah'}=\sqrt{(15)^2+2(-9.8)(5.74)}=10.6 m/s$

And therefore, the momentum of the ball here is

$p'=mv'=(0.1)(10.6)=1.06 kg m/s$

Learn more about momentum:

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#LearnwithBrainly

7 0

3 years ago

A bird can fly 25 km/hr. How long does it take to fly 10 km?

Svetradugi [14.3K]

Answer:

25 minutes

Explanation:

4 0

3 years ago

Help please ! USA test prep

Illusion [34]

C) 2 ohms I think don't fully trust my answer I suck at math but I think it's c

3 0

3 years ago

•• Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that

Y_Kistochka [10]

Answer:

$v=0.57\frac{m}{s}$

$a_c=10.83\frac{m}{s^2}$

Explanation:

We have an uniform circular motion, therefore, the pebble’s speed is given by the distance traveled in a revolution $(2\pi r)$ and the period (T), since this is the time pebble’s takes to complete a revolution:

$v=\frac{2\pi r}{T}$

The period is inversely proportional to the frequency:

$T=\frac{1}{f}$

So, we have:

$v=\frac{2\pi r}{\frac{1}{f}}\\v=2\pi rf\\$

Recall that the radius is the half of the diameter and one revolution per is equal to one Hz:

$v=2\pi (30*10^{-2}m)(3Hz)\\v=0.57\frac{m}{s}$

The centripetal acceleration is defined as:

$a_c=\frac{v^2}{r}\\a_c=\frac{(0.57\frac{m}{s})^2}{30*10^{-2}m}\\\\a_c=10.83\frac{m}{s^2}$

6 0

3 years ago

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 5.80 mm behind the apert

Helga [31]

Explanation:

Whenever the light passes through hole or slit then it tends to bend that is actually a diffraction. It will then made the interference pattern of light and dark bands that due to constructive and destructive interference.

So by using the equation of diffraction,

dsinA = nL

SinA is a geometric component it can be written as,

$d\frac{y}{x} = nL$

∵ x is the distance from screen.

∵ y is the half of the width of central maximum.

now by putting the values in mm,

$d\frac{17}{5.80} = 0.000633$

d = 2.15× $10^{-4}$ mm

7 0

3 years ago