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olga nikolaevna [1]
3 years ago
8

Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p

uck begins to move to the moment it hits the spring. Use 0.253 m for the displacement of the puck along the ramp and 9.80 m/s2 for the acceleration due to gravity. Assume that the mass of the puck is 0.180 kg. Express your answer using SI units to three significant figures.
Physics
1 answer:
jekas [21]3 years ago
7 0

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

You might be interested in
The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand
AlexFokin [52]

Answer:

8.67791\times 10^{25}\ kg

0.34589\ m/s^2

0.07903\ m/s^2

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

r_m = Radius of Miranda = 236 km

m = Mass of Miranda = 6.6\times 10^{19}\ kg

Acceleration due to gravity is given by

g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg

The mass of Uranus is 8.67791\times 10^{25}\ kg

Acceleration is given by

a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2

Miranda's acceleration due to its orbital motion about Uranus is 0.34589\ m/s^2

On Miranda

g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2

Acceleration due to Miranda's gravity at the surface of Miranda is 0.07903\ m/s^2

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0
3 years ago
Please help as fast you can
Ilya [14]

Answer:

d a

Explanation:

7 0
3 years ago
A horizontal force of magnitude 356 N is employed to push a 78.0-kg block a distance of 18.6 m on a rough horizontal surface. If
icang [17]

Answer:

6621.6 Joule

0.46525

Explanation:

F = Force = 356 N

s = Displacement = 18.6 m

m = Mass of block = 78 kg

Work done

W=F\times s\\\Rightarrow W=356\times 18.6\\\Rightarrow W=6621.6\ J

The work done by the horizontal force is 6621.6 Joule

F=\mu mg\\\Rightarrow \mu=\frac{F}{mg}\\\Rightarrow \mu=\frac{356}{78\times 9.81}\\\Rightarrow \mu=0.46525

The coefficient of friction between the block and the rough surface is 0.46525

5 0
3 years ago
Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2
Mrrafil [7]

Answer:

The girl exerts more pressure.

Explanation:

Pressure can be defined as the force exerted normally or perpendicularly per unit area.

i.e P = F/A

<u>Girls</u>

Area of the heel = 1cm² = 10^(-4) m²

Force = mg = 50 × 10 = 500N

Pressure =

\frac{500}{10 ^{ - 4} }

= 5 \times  {10}^{6}

<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u>\frac{40000}{2.5 \times  {10}^{ - 2} }</u>

<u>= 1.6 \times  {10}^{6}</u>

5 0
3 years ago
Blocks A and B of unknown masses m1 and m2, respectively, are set up on an inclined plane as shown. Block A is attached to block
Korvikt [17]

Newton's second law we can find that the correct answer is:  

 E)  It cannot be determiner whick block has more masses from the information provided

Newton's second law establishes the relationship between force, mass, and acceleration of a body. Since force and acceleration are vector quantities, their components must be added on each axis

For this problem we have two bodies, let's write Newton's second law for the body B, we assume that the body B descends

            W_b - T = m_b a

            W_b  = m_b g

            m_b - T = m_b a

Where W_b is the weight of block B, T the tension of the string, mb the mass of block b and the acceleration

Now let's find the relation for block A

let's set a datum with the x axis parallel to the ramp

           T - Wₓ = mₐ a

           sin θ = Wₓ / W

            Wₓ = Wₐ sin θ

             Wₐ = mₐ g

Where Wₓ is the component of the weight, Wₐ the weight of the body A and θ the angle of the plane

Let's write our system of equations

           m_b g - T = m_b a

           T - mₐ g sin θ = mₐ a

let's add the equations

            g (m_b - mₐ sin θ) = (m_b + mₐ) a

            a =   \frac{m_b - m_a \ sin  \ \theta}{m_b+m_a} \ g

Let's analyze this expression

  • The numerator is positive the body B descends, this occurs when

          m_b - mₐ sin θ > 0

           

  • The numerator is negative, body B rises

           m_b - mₐ  sin θ <0

We can observe that the acceleration is positive or negative depending on the relation of the masses and the angle of the plane.

In conclusion using Newton's second law we find that the correct answer is  

 E )   It cannot be determiner whick block has more masses from the information provided

learn more about Newton's second law here:

brainly.com/question/9099891

8 0
3 years ago
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