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2 years ago

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A 21200 kg sailboat experiences an eastward force 42700 N due to the tide pushing its hull while the wind pushes the sails with

a force of 85000 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? Answer in units of m/s 2 .

1 answer:

My name is Ann [436]2 years ago

7 0

Answer:

$2.95 m/s^{2}$

Explanation:

The resultant force F

$F=\sqrt {F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos135^{o}}$ where $F_{1}$ is eastward force, $F_{2}$ is force directed towards the North

$F=\sqrt {42700^{2}+85000^{2}+(2*42700*85000)cos135^{o}}=62573.17217 N$

F=62573.2 N

The magnitude of acceleration of sailboat is given by

$a=\frac {F}{m}=\frac {62573.2}{21200}=2.95 m/s^{2}$

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Which of the following planets helped astronomers locate another planet?

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Answer: Pluto,Mercury

Explanation:

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2 years ago

Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th

maria [59]

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.

c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9

d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4

e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

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2 years ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

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Answer:

$4.1\cdot 10^8 N$

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

$g=9.8 m/s^2$ is the gravitational acceleration

$h=11,000 m$ is the depth

Substituting,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

And so, the inward force exerted on it is

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

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2 years ago

Read 2 more answers

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has

NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, $\dfrac{m}{t}=30\ kg/s$

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

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2 years ago

You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera

Nina [5.8K]

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁ = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = $\frac{2*4186*5}{0.5*25}$ = 3348.8J/kg C≅ 3349J/kg C

4 0

2 years ago