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slamgirl [31]
3 years ago
11

A 20600 kg sailboat experiences an eastward force 17700 N due to the tide pushing its hullwhile the wind pushes the sails with a

force of 60000 N directed toward the northwest (45◦ westward of North or 45◦ northward of West). What is the magnitude of the resultant acceleration of the sailboat?
Physics
1 answer:
gavmur [86]3 years ago
3 0

Answer:

2.23 m/s²

Explanation:

mass of sail boat, m = 20600 kg

Force along east, F1 = 17700 N

Force along north west, F2 = 60000 N

write the forces in vector form

\overrightarrow{F_{1}}=17700 \widehat{i}

\overrightarrow{F_{2}}=60000\left ( - Cos45 \widehat{i}+Sin45\widehat{j}\right )

\overrightarrow{F_{2}}=-42426.41\widehat{i}+42426.41\widehat{j}

Net force acting on the boat

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=17573.6\widehat{i}+42426.41\widehat{j}

The magnitude of net force is given by

F=\sqrt{17573.6^{2}+42426.41^{2}}

F = 45922 N

Acceleration = force / mass

a = 45922 / 20600

a = 2.23 m/s²

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Time(t)=2 seconds

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Distance=(6 x 2 x 2)/2

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A hot-air balloon has a volume of 440 × 10^3 ^3. Calculate the buoyant force that the surrounding cold air exerts on the balloon
siniylev [52]

Answer:

Explanation:

Given that,

The volume of the balloon is

V = 440 × 10³ m³

Buoyant force F?

Given the density of the surrounding to be 2.58 kg/m³

ρ = 2.58 kg/m³

The buoyant force is the weight of water displaced and it is calculated using

F_b = ρVg

Where

F_b is buoyant force

ρ is density

V is the volume of the liquid displace.

g is the acceleration due to gravity

Then,

F_b = ρVg

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3 0
3 years ago
A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
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