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givi [52]
3 years ago
7

A solid ball is rolling along a horizontal surface at 3.7 m/s when it encounters an upward

Physics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

h = 0.697 [m]  

Explanation:

To solve this problem we must use the energy conservation theorem, where it tells us that kinetic energy is converted to potential energy or vice versa.

E_{kinet}=E_{pot}

where:

Ekinet = kinetic energy [J]

Epot = potential energy [J]

\frac{1}{2}*m*v^{2}=m*g*h\\0.5*(3.7)^{2} =9.81*h\\h =(6.845)/9.81\\h = 0.697 [m]

You might be interested in
Mars has a mass of about 6.58 × 1023 kg, and its moon Phobos has a mass of about 9.3 × 1015 kg. If the magnitude of the gravitat
NISA [10]

This looks complicated, but it's actually not too tough.

The formula for the gravitational force between two objects is

              Force = G  (one mass) (other mass) / (distance²) .

The question GAVE us all of those numbers except the distance.
All we have to do is pluggum in, massage it around, and find
the distance. 

Force  =         4.18 x 10¹⁵     N
  G  =             6.673 x 10⁻¹¹  N·m²/kg²
One mass =   6.58 x 10²³     kg
Other mass = 9.3 x 10¹⁵       kg   .

The only tricky thing about this is gonna be the arithmetic ...
keeping all the exponents straight.

Take the formula for the gravitational force and plug in
everything we know:

Force = (G) · (one mass) · (other mass) / (distance²) 

4.18x10¹⁵N = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (distance²).

Multiply each side by  (distance²):

(distance²)·(4.18x10¹⁵N) = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) 

Divide each side by  (4.18 x 10¹⁵ N) :

(distance²)=(6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (4.18x10¹⁵N)

That's the end of the Physics and Algebra.  The only thing left is Arithmetic.
We have to simplify that whole ugly thing on the right side of the equation,
and then take the square root of each side.

When I crunch down the right side of that equation, I get

           (distance²)  =  9.769 x 10¹³  m²

and when I take the square root of each side, I get

             distance  =  9.884 x 10⁶ meters .   **

You should check my Arithmetic.   **
(Pause occasionally to let your calculator cool off.)


BY THE WAY ... 
That "distance" in the equation for gravitational force is the distance
between the CENTERS of the two objects. 
This doesn't make much difference for Phobos, because Phobos isn't
much bigger than a big sweet potato.  But it does make a difference for
Mars. 
The 'distance' we find with all of this nonsense is NOT the distance
between Phobos and the surface of Mars.  It's the distance between 
Phobos and the CENTER of Mars, so it includes the planet's radius.   


** Consulting online resources between Floogle and Flickerpedia,
I found that the orbital distance of Phobos from Mars varies between
9,234 km and 9,517 km.  Add the planet's radius to these, and I'm
beginning to feel confidence in the results of my back-of-the-napkin
calculation.  But you should still check my Arithmetic.

5 0
3 years ago
A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simu
natta225 [31]
 <span>a) M is the big one 
m is the little one 

v is the speed of each of them when they impact 

V = speed of mii after collision 

Conservation of momentum 

Mv – mv = mV 

Conservation of energy 

1/2Mv^2 + 1/2mv^2 = 1/2 mV^2 

This pair simplify to give 

M = 3m 

V = 2v 

So m = 0.21kg 

and h = 4 . 2.7 = 10.8 m</span>Source(s):<span>Old teacher</span>
8 0
3 years ago
. What is the height in meters of a 5'4" person?​
Illusion [34]

Answer:

1.62 meters

Explanation:

For an approximate result, divide the feet (height) by 3.3 to get your approximate meters (height).

5 0
3 years ago
Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge
Vanyuwa [196]

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k k \frac{ q_{1}  \  q_{2} }{ r^{2} }

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = F_{bc x}

Y axis  

       F_{y}Fy = F_{ab} - F_{bc y}

let's find the magnitude of each force

     F_{ab} = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

   

      F_{bc} = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + F_{y} j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = \sqrt{ F_{x}^2 + F_{y}^2 }

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= \frac{ F_{y}  }{  F_{x} }

       

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

3 0
3 years ago
Which is an example of radiation?
oksano4ka [1.4K]
I believe The snowman should be the answer
5 0
3 years ago
Read 2 more answers
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