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Naddik [55]
1 year ago
5

Scenario 3Starting at rest, a 3 kg ball is dropped from the side of a bridge and strikes the ground below at 35m/s. What is the

height of the bridge?
Physics
1 answer:
Nonamiya [84]1 year ago
4 0

   The ball's gravitational potential energy is converted into kinetic energy as it falls toward the ground.

<h3>How can the height of a dropped ball be determined?</h3>

    Y = 1/2 g t 2, where y is the height above the ground, g = 9.8 m/s2, and t = 1.3 s, is the formula for problems like these. Any freely falling body with an initial velocity of zero meters per second can use this formula. figuring out how much y is.

   A ball drops from the top of a building and picks up speed as it descends. Its speed is increasing by 10 m/s every second. What we refer to as motion with constant acceleration is, for example, a ball falling due to gravity.

    The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight downhill motion, which has a speed of 17.1 m/s. Take note of the rising positive y direction in the above graphic.

To Learn more About potential energy, Refer:

brainly.com/question/14427111

#SPJ10

You might be interested in
Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+
babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• \mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j with 0\le x\le2 and 0\le y\le6-x;

• \mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k with 0\le u\le2 and 0\le v\le\frac\pi2;

• \mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k with 0\le y\le 6 and 0\le z\le2;

• \mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k with 0\le u\le2 and 0\le v\le\frac\pi2; and

• \mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k with 0\le u\le\frac\pi2 and 0\le y\le6-2\cos u.

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k

\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j

\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i

\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j

\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0

\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du

\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8

\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz

=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0

\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du

=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi

\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du

=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV

where <em>R</em> is the interior of <em>S</em>. We have

\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7

The integral is easily computed in cylindrical coordinates:

\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2

\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3

as expected.

4 0
3 years ago
At what condition does a body becomes weightless at the equator?
ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

4 0
3 years ago
What is the velocity (m/s) after 5 seconds of fall
gregori [183]
5 second fall starting at 0 m/s
ball strikes ground at a speed = 49 meters per second.
3 0
3 years ago
An electromagnetic wave of wavelength
Ivanshal [37]

Answer:

4.01\cdot 10^{-7} m

Explanation:

When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.

In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength in a vacuum (air is a good approximation of vacuum)

n is the refractive index of the medium

In this problem:

\lambda_0 = 5.89\cdot 10^{-7} m is the original wavelength of the wave

n = 1.47 is the index of refraction of corn oil

Therefore, the wavelength of the electromagnetic wave in corn oil is:

\lambda=\frac{5.89\cdot 10^{-7}}{1.47}=4.01\cdot 10^{-7} m

8 0
3 years ago
The position-time graph for a bug crawling along a line is shown in item 4 below. Determine whether the velocity is positive, ne
Naddika [18.5K]

Answer: The velocity at different marked time points are given as

t1 = -

t2 = +

t3 = +

t4 = -

t5 = 0

Explanation:

The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal

axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.

When the tangent of the line is parallel to the horizontal axis, the velocity is 0.

From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below

t1 = -

t2 = +

t3 = +

t4 = -

t5 = 0

QED!

5 0
3 years ago
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