Ask question

Ask question

All categories

antoniya [11.8K]

3 years ago

6

A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

inal velocity. find the time it takes the boat to travel this distance.

1 answer:

sasho [114]3 years ago

8 0

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

You might be interested in

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

How do two sublevels of the same principal energy level differ from each other

Colt1911 [192]

Each energy sublevel corresponds to an orbital of a different shape.

Explanation:

Two sublevels of the same principal energy level differs from each other if the sublevels corrresponds to an orbital of a different shape.

The principal quantum number of an atom represents the main energy level in which the orbital is located or the distance of an orbital from the nucleus. It takes values of n = 1,2,3,4 et.c
The secondary quantum number gives the shape of the orbitals in subshells accommodating electrons.
The number of possible shapes is limited by the principal quantum numbers.

Take for example, Carbon:

1s² 2s² 2p²

The second energy level is 2 but with two different sublevels of s and p. They have different shapes. S is spherical and P is dumb-bell shaped .

Learn more:

Quantum number brainly.com/question/9288609

#learnwithBrainly

7 0

3 years ago

Ira Lisetskai [31]

The answer is magnet away from the coil

3 0

3 years ago

Read 2 more answers

Which statement is true regarding a distance vs. time graph? (1 point)

Nadusha1986 [10]

The statement that is true regarding a distance vs. time graph is option A: The graph should show distance on the vertical axis.

<h3>Where is the plot of distance?</h3>

How far an object has come in a certain amount of time is displayed on a distance-time graph. Time is represented on the X-axis and Distance is plotted on the Y-axis (left) (bottom).

On a distance-time graph, an object's motion is indicated by a sloping line. The slope or gradient of the line in a distance-time graph is equal to the object's speed. The object is travelling more quickly the steeper the line is (and the bigger the gradient).

Note that the distance-time graph shows the relationship between distance and time by plotting distance on the y-axis and time on the x-axis.

Learn more about distance vs. time graph from

brainly.com/question/16825120
#SPJ1

6 0

1 year ago

Read 2 more answers

What happens to the movement of particles in a liquid when it is heated?

Yanka [14]

Answer:

When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increases. The motion and spacing of the particles determines the state of matter of the substance. The end result of increased molecular motion is that the object expands and takes up more space.

Explanation:

3 0

3 years ago

Read 2 more answers