A horse was grazing at the top of a small hill. The horse got thirsty and ran toward a

A wave can be best defined as <br><br> P.S pls help

Sidana [21]

Answer:

It should be A. Disturbance that travels through a medium or space, transmitting energy from one point to another.

I hope this helped you :)

4 0

2 years ago

Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the

makvit [3.9K]

Answer:

$\mu_k=0.101$

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

$\dfrac{1}{2}kx^2=\mu_k mgh$

$\mu_k=\dfrac{kx^2}{2mgh}$

$\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}$

$\mu_k=0.101$

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

Listed following is a set of statements describing individual stars or characteristics of stars. Match these to the appropriate

Oksana_A [137]

Answer:

Red giant or super giant → very cool but very luminous

→ found in the upper right of the H-R diagram.

Main sequence →The majority of stars in our galaxy

→ Sun, for example

→ a very hot and very luminous star

White dwarfs → very hot but very dim

→ not much larger in radius than earth

Explanation:

Giant:

When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars.That's why they are very cool. Giant stars have the larger radius and luminosity then the main sequence stars.

Main Sequence:

Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. That is why they are very hot and luminous. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.

White dwarfs:

When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.

6 0

3 years ago

If you begin with 40 grams of a radioactive isotope and end with 10 grams, how many half-lifes of the radioactive isotope have p

Rom4ik [11]

30 grams of radioactive isotope have passed.

8 0

3 years ago