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son4ous [18]
4 years ago
6

Mr. Phillips' car iso parked on a steep hill

Physics
1 answer:
Wewaii [24]4 years ago
7 0
? There is no question
You might be interested in
A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
balu736 [363]

Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

-- The buoyant force is precisely the missing <em>30N</em> .

--  In order to calculate the density of the frewium sample, we need to know
its mass and its volume.  Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

   Weight = (mass) x (gravity)
   185N = (mass) x (9.81 m/s²)
   Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u> 

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³.  So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

           30N = (mass of the displaced water) x (9.81 m/s²)

           Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

           Volume of displaced water = <u>3,058 cm³</u>

Finally, density of the frewium sample = (mass)/(volume)

      Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)

================================================

I'm thinking that this must  be the hard way to do it,
because I noticed that

       (weight in air) / (buoyant force) =  185N / 30N = <u>6.1666...</u>

So apparently . . .

        (density of a sample) / (density of water) =

                                  (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.


3 0
3 years ago
Definition for compression and rarefraction
s344n2d4d5 [400]
Compression and rarefaction are two phenomenon occurs in longitudunal wave!

when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !
6 0
3 years ago
Read 2 more answers
An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t
erik [133]

Answer:

Bounce 1 ,  pass 3,   emb2

Explanation:

(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle.  So it is  Bounce1, pass3, emb2.  

6 0
3 years ago
Draw the following patterns as they would appear when viewed through a compound microscope.:
goblinko [34]
2 j x b
H c e c hope this helppppsssssssss
3 0
3 years ago
Read 2 more answers
A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist
Paha777 [63]

Answer:

Part a)

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

\frac{1}{2}mv^2 = mgh'

now we know that

I = \frac{2}{5}mr^2

\omega = \frac{v}{r}

so we have

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh

\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh

\frac{7}{10}mv^2 = mgh

\frac{1}{2}mv^2 = \frac{10}{14}mgh

so the height on the smooth side is given as

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

6 0
4 years ago
Read 2 more answers
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