Describe briefly how you would determine the density an irregular object (stone)

makvit [3.9K]

The Earth revolves (orbits) around the Sun in one year. The Earth's rotation axis is tilted relative to the plane of its orbit around the Sun. This tilt of the Earth is responsible for the seasons as the Earth orbits the Sun. The Sun provides energy that sustains all life on Earth.

4 0

3 years ago

Guys helppppp pleaseeeeeeeee

yan [13]

2 pencils, because that would equal left grams then the others

4 0

3 years ago

You walk 100m due north. You then turn and walk 55m due east. You then make another turn and walk 12m due south. What is the res

lord [1]

Answer:

Explanation:

Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

$A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55$

Add them all together to get the x component of the resultant vector, V:

$V_x=55$

Do the same to find the y components of the part of this journey:

$A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12$

Add them together to get the y component of the resultant vector, V:

$V_y=88$

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.

We find the final magnitude:

$V_{mag}=\sqrt{55^2+88^2}$ and, rounding to 2 sig dig's as needed:

$V_{mag}=$ 1.0 × 10² m; now for the direction:

$\theta=tan^{-1}(\frac{88}{55})=$ 58°

7 0

2 years ago

Then it is called

DiKsa [7]

Answer:

It is called force of friction

Explanation:

The force of friction is a force that acts between two objects whose surfaces are in contact with each other.

Consider the typical case of an object sliding along a certain surface. There are two types of frictions:

- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:

$F_f = F$

However, this force of static friction has a maximum value, which is given by

$F_{max} = \mu_s N$

where

$\mu_s$ is the coefficient of static friction

N is the normal reaction exerted by the surface on the object

So, when $F$ becomes greater than $F_{max}$ , the static friction is no longer able to balance the push applied, and the object will start sliding forward.

- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by

$F_f = \mu_k N$

where

$\mu_k$ is the coefficient of kinetic friction, and its value is generally smaller than $\mu_s$ . The direction of this force is also opposite to the direction of motion of the object.

8 0

3 years ago

A projectile of mass m is launched with an initial velocity vector v i making an angle θ with the horizontal as shown below. The

sergeinik [125]

Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be:
L = R * m * vi * cos(90 - theta) 

cos(90 - theta) is just sin(theta) 
and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g 

so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g 

We can combine the two vi terms and get: 

L = vi^3 * m * sin(theta) * sin(2*theta) / g 

What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. 

Now, for the first part... 

There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. 

Okay, so let's get back to what we know: 

L = d * m * v * cos(phi) 

where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... 

m is still m. 
v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) 
d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g 

That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. 

The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: 

h = 1/2 * (vi * sin(theta))^2 / g 

So, there's the rise. So, our *slope* is rise/run, so 

slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] 

The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: 

slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] 

(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). 

Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... 

slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 

Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... 

alpha = arctan( tan(theta) / 4 ) 

Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha 

so, phi = 180 - arctan( tan(theta) / 4 ) 

Now, we go back to our original formula and plug it ALL in... 

L = d * m * v * cos(phi) 

becomes... 

L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] 

Now, cos(180 - something) = cos(something), so we can simplify a little bit... 

L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ]

3 0

3 years ago

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