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attashe74 [19]
3 years ago
10

A girl is bouncing on a trampoline where is her gravitational potential energy a maximum and where is her kinetic energy maximum

Physics
1 answer:
Stella [2.4K]3 years ago
7 0

Answer:

When you jump down, your kinetic is converted to potential energy of the stretched trampoline. The trampoline's potential energy is converted into kinetic energy, which is transferred to you, making you bounce up. At the top of your jump, all your kinetic energy has been converted into potential energy. Right before you hit the trampoline, all of your potential energy has  been converted back into kinetic energy. As you jump up and down your kinetic energy increases and decrease.

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a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100
Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

       

let's reduce the magnitudes to the SI system

         r = 10 cm = 0.10 m

         m = 10 g = 0.010 kg

         A = 100 cm² = 0.01 m²

the definition of density is

          ρ = m / V

the volume of a sphere

         V = \frac{4}{3} \ \pi r^{3}

          V = \frac{4}{3} π 0.1³

          V = 4.189 10⁻³ m³

let's calculate the density of the ball

           ρ = \frac{0.010}{4.189 \ 10^{-3} }

           ρ = 2.387 kg / m³

the tabulated density of water is

         ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

            B - W = 0

            B = W

             

where B is the thrust that is given by Archimedes' principle

           ρ_liquid  g V_submerged = m g

           V_submerged = m / ρ_liquid

we calculate

            V _submerged = 0.10 9.8 / 997

             V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

           V = A h

            h = V / A

let's calculate

            h = 9.83 10⁻⁴ / 0.01

            h = 0.0983  m

this is equal to h = 9.83 cm

8 0
3 years ago
If you want quick access to large amount of data, you would use a
RSB [31]
You would use d. table

6 0
3 years ago
I need help in my physics class and show me how it’s done
Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)

Where | A | is the magnitude of the vector and \alpha is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle \alpha.

Vector A is in the 4th quadrant

So:

A_x = 6\\\\A_y = -6.5

So:

|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846

So:

Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)

\alpha = -47.28 ° +360° = 313 °

\alpha = 313 °

Option 4.

4 0
3 years ago
What is the farthest distance parallaxes can be used to measure star distances from Earth?
JulsSmile [24]
Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.
8 0
3 years ago
A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

5 0
3 years ago
Read 2 more answers
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