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3 years ago

Determine the time needed for the load at b to attain a speed of 10 m/s, starting from rest.

1 answer:

5 0

Unfortunately, I can't give a definite answer because I only know two information: the initial velocity which is 0 because it is at rest, and the final velocity which is 10 m/s. To solve this, I should know one other information to determine time. It could be the constant acceleration. The useful equation to be used is:

a = (vf - v₀)/t

So, assuming that the object was dropped from a certain height, the acceleration is due to gravity which is equal to 9.81 m/s². Then,

9.81 = (10 - 0)/t
t = 1.02 seconds

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A scuba diver and her gear displace a volume of 67.0 l and have a total mass of 64.0 kg. (a) what is the buoyant force on the di

slavikrds [6]

Buoyant force is the force that is a result from the pressure exerted by a fluid on the object. We calculate this value by using the Archimedes principle where it says that the upward buoyant force that is being exerted to a body that is immersed in the fluid is equal to the fluid's weight that the object has displaced. Buoyant force always acts opposing the direction of weight. We calculate as follows:

Fb = W
Fb = mass (acceleration due to gravity)
Fb = 64.0 kg ( 9.81 m/s^2)
Fb = 627.84 kg m/s^2

Therefore, the buoyant force that is exerted on the diver in the sea water would be 627.84 N

4 0

3 years ago

3. The soccer kicker kicks a 2 kg football with a force of 68 N. How fast will the ball accelerate down the field?

Katen [24]

Answer:

$\boxed {\boxed {\sf m= 2 \ kg , \\a= 34 \ m/s^2, \\\F= 68 \ N}}$

Explanation:

The formula for force is:

$F=m*a$

If we rearrange the formula to solve for a (acceleration), the formula becomes

$\frac{F}{m} =a$

The force is 68 Newtons. Let's convert the units to make the problem easier later on. 1 N is equal to 1 kg*m/s², so the force of 68 N is equal to 68 kg*m/s².

The mass is 2 kilograms.

$F=68 \ kg*m/s^2 \\m= 2\ kg$

Substitute the values into the formula.

$\frac{68 \ kg*m/s^2}{2 \ kg} =a$

Divide. Note that the kilograms will cancel each other out (hence why we changed the units).

$\frac{68 \ m/s^2}{2}=a$

$34 \ m/s^2=a$

The acceleration is 34 meters per second squared.

6 0

3 years ago

Imagine an infinite earth with a hole dripped through it. You fall in and accelerate at g~10m/s/s. How long until you reach the

Soloha48 [4]

An object with non-zero mass (even negligible mass is non-zero) will never reach the speed of light. Due to relativistic effects, each "unit" of acceleration becomes less effective at increasing your velocity (relative to some other object, of course) as your relative velocity approaches the speed of light.

And even if there was a way, If you would accelerate to the 99,99% of the speed light in just 1 second, you would experience a G-force of aprox. 30,600,000 g's which is enough to kill you in a few seconds

6 0

3 years ago

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 f 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 S 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 R is the interior of S. 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

A player kick the soccer ball from ground level and send it flying at an angle of 30° at a speed of 26M/S. What is the maximum h

icang [17]

The answer would be 2.63. Your welcome. This has been changed to the correct answer.

7 0

3 years ago