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3 years ago

Why does a ship need an anchor

2 answers:

6 0

A ship needs an anchor because if a ship stop its just going to float around the ocean, but if the ship has the anchor then the ship will stay in place.

Len [333]3 years ago

5 0

So there is extra wait that pulls the boat to stop.

Can i get brainliest I need it

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Write the following number in scientific notation: 658000

Mice21 [21]

6.58 x 105 is the answer

8 0

3 years ago

A wheelbarrow is experiencing two forces. One is pushing 5 N to the east and the other 5 N in the exact opposite direction. What

mezya [45]

Assuming no other forces are acting on the wheelbarrow, it must be stationary.

4 0

3 years ago

Read 2 more answers

A ball of mass 4.5 kg moving with speed of 2.2 m/s in the +x-direction hits a wall and bounces back with the same speed in the x

Levart [38]

Answer:

The change of the momentum of the ball is $-19.8\, \frac{mkg}{s}$

Explanation:

We should find $\varDelta\overrightarrow{p}=\overrightarrow{p_{f}}-\overrightarrow{p_{i}}$ (1)with $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum. Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (1):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (2)

It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall $\overrightarrow{v_{i}}=+2.2\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-2.2\, \frac{m}{s}$ so (2) becomes:

$\varDelta\overrightarrow{p}=m(-2.2- (+2.2))=-(4.5)(4.4)$

$\varDelta\overrightarrow{p}=-19.8\, \frac{mkg}{s}$

8 0

3 years ago

Find the following answer based on the image.

saul85 [17]

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

5 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago