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3 years ago

What happens when an object with a higher density is placed in a container with a lower density liquid?

1 answer:

8 0

The object with the higher density sinks to the bottom and the lower density stays at the top.They don't mix together either because, of the type of density.

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A sort of "projectile launcher" is shown below. A large current moves in a closed loop composed of fixed rails, a power supply,

Zarrin [17]

Answer:

wallah i don't understand anything with my stoopid brain

Explanation:

3 0

3 years ago

For work to be done on the object the object has to

Nata [24]

Gain energy

good luck!

3 0

3 years ago

Read 2 more answers

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

4 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago

A crate is to be pulled across the floor. The crate will experience a force of friction of 34 N. You pull on the crate with a fo

Jobisdone [24]

The net force would be 75N - 34N = 41N

The crates rate of acceleration would be 41N=8.5*a so a = 4.8m/s^2

6 0

3 years ago