Ask question

Ask question

All categories

3 years ago

8

A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3

ft. above home plate when hit by the batter, what is the ball's increase in potential energy?

1 answer:

den301095 [7]3 years ago

5 0

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

$h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m$

$h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m$

$m = 148g*\frac{1kg}{1000g} = 0.148kg$

$g = 9.8 \frac{m}{s^2}$

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

$U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J$

The increase in potential energy of the ball is 115.82 J

You might be interested in

monitta

Answer:

See explanation below.

Explanation:

The internal energy (kinetic energy) of the milk particles increases thus warming the milk up, while the internal energy of the coffee particles reduce their initial average speed as they are imparting energy to the slow milk particles.

7 0

3 years ago

A wave of infrared light has a speed of 6m/s and a wave length of 12 m. What is the frequency of this wave?

erastovalidia [21]

Frequency = speed / wavelength

(6 m/s) / (12 m) = 0.5 Hz.

That's not infrared light.
Infrared light waves move about 50 million times faster than that, and they're only about 0.00000007 as long as that.

6 0

4 years ago

I’m stuck on this problem, help is very appreciated!!

yaroslaw [1]

Explanation:

Use the equation F=m×a

F= 1500 N

M= 500kg

a= F/M

a= 1500÷ 500

a= 3m/s^2

8 0

3 years ago

Read 2 more answers

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

Romeo (81.0 kg) entertains Juliet (53.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m awa

den301095 [7]

Answer:

0.64 m

Explanation:

The first thing is calculate the center of mass of the system.

$X_{cm}= \sum_{n=1}^{n}\frac{X_n\times M_n}{M_n}$

now multiplying every coordinate x by the mass of each object (romeo, juliet and the boat) and dividing all by the total mass taking by reference the position of juliet.

$X_{cm}=\frac{53\times0 +81\times2.7+79\times1.35}{53+81+79}$

X_cm = 1.4589 m

When the forces involved are internals, the center of mass don't change

After the movement the center of mass remains in the same distance from the shore, but change relative to the rear of the boat.

$X_{cm}=\frac{79\times1.35+(81+53)\times2.7}{53+81+79}$

X_cm= 2.10 m

this displacement is how the boat move toward the shore.

2.10-1.46= 0.64 m

5 0

3 years ago