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katrin2010 [14]
3 years ago
7

A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

Physics
1 answer:
pogonyaev3 years ago
5 0

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

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A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to
loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of V_{x} (horizontal component of velocity) and V_{y} (vertical component of velocity)

We have

V_{y} = Velocity in South direction = 18m/s

m_{y} = Mass of piece in South direction = m

V_{x} = Velocity in West direction = 18m/s

m_{x} = Mass of piece in West direction = m

V_{ry} = Velocity of Third Piece in North direction = unknown

V_{rx} = Velocity of Third Piece in East direction = unknown

m = Mass of third piece = 2m

mV_{ry}  = m_{y} V_{y} \\ 2mV_{ry}  = m18\\ V_{ry} = 9

mV_{rx}  = m_{x} V_{x} \\ 2mV_{rx}  = m18\\ V_{rx} = 9

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

Hypotenuse^{2}  = Height^{2}  + Base^{2} \\ V_{3}^2  = 9^{2} + 9^{2} \\ V = 12.728 m/s

V = 13m/s

6 0
3 years ago
A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?
Anna35 [415]

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2       multiply both sides by 2

8.5 = 4.095 t^2                  Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2                       Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

6 0
3 years ago
When traveling on narrow mountain roads _______________. A. honk your horn if you cannot see at least 200 ft ahead B. expect oth
yarga [219]

Answer:

The correct option is;

A. honk your horn if you cannot see at least 200 ft ahead

Explanation:

According the California Driver Handbook on Safe Driving Practices, it is required of the driver driving on a narrow mountain road without clear visualization of what is 200 ft ahead of her or him to honk the horn of the vehicle.

The sounding of the horn will alert those ahead of the driver of the possible danger due to her or his oncoming vehicle so that they (those ahead of the driver's oncoming vehicle) can react appropriately.

6 0
3 years ago
La) What is meant by the term Basic Quantities.​
castortr0y [4]

Answer:

A basic quantity is basically the physical quantity that can not be defined in terms of other quantities.

Explanation:

A basic quantity is basically the physical quantity that can not be defined in terms of other quantities.

Some of the names of the basic quantities include:

  1. Mass, denoted by the symbol 'm', with S.I. unit 'kg'
  2. Length, denoted by symbol 'l', with S.I. unit 'm'
  3. Time, denoted by symbol 't', with S.I. unit 's'
  4. Current, denoted by 'I', with S.I. unit 's' 'A'
  5. Temperature, denoted by 'T', with S.I. unit 'K'
  6. Amount of substance, denoted by 'n', with S.I. unit 'mol'
  7. Luminous Intensity, denoted by 'Iv', with S.I. unit 'cd'

  • A basic quantity is chosen arbitrarily.
5 0
2 years ago
Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v
vladimir1956 [14]

Answer:

P(X\ge 1) = 0.9502

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

\lambda = density

\lambda = \frac{3}{10}

\lambda = 0.3

T = the light years

T = 10

Calculating P(X \ge 1)

In probability:

P(X \ge 1) = 1 - P(X = 0)

Calculating P(X=0)

Substitute 0 for k and the values for \lambda and T in

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}

P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}

P(X=0) = (3)^0 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-3}

P(X=0) = e^{-3}

P(X=0) = 0.04979

Substitute 0.04979 for P(X=0) in P(X \ge 1) = 1 - P(X = 0)

P(X\ge 1) = 1 - 0.04979

P(X\ge 1) = 0.95021

P(X\ge 1) = 0.9502 ---  approximated

<em>Hence, the required probability is 0.9502</em>

3 0
3 years ago
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