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Naya [18.7K]
2 years ago
9

Alicia can row 6 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 3 miles/hour faster

than she rows upstream. Find Alicia’s rowing rate each way. Round your answers to the nearest tenth, if necessary.
Physics
2 answers:
m_a_m_a [10]2 years ago
4 0
Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
olga_2 [115]2 years ago
4 0

Full...Solving Rational Equations Quiz part 1.

1.c. n^2-6/n^2-2 ; n = +/- sqrt5, n= +/- sqrt2

2.B.  4a/7b^2 , a = 0, b = 0

3.C. (x-4)^2/(x+3)(x+1) ; x= -4,-3,-2,-1,4

4.B. (x+1)(x-1)(x^2+1)

5.A. 7a-49/(a-8)(a+8)

6.A. 21a-28/(A-6)(a+8)

7.C. 4x/3x^2+10x+3

8.C. 3x^2(y+4)/7y

9.D. -11/3

10.D. 14

11. D. 9 mi/h downstream, 6 mi/h upstream

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Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written asx(t)=Asin(\omega t+\phi)

'A' is the amplitude = 6 inches (given)

\omega =\sqrt{\frac{k}{m}} is the natural frequency of the system

At equilibrium we have

mg=kx\\\\k=\frac{mg}{x}

Applying values we get

k=40 lb/ft

thus natural frequency equals

\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}

Thus the equation of motion becomes

x(t)=6sin(8t+\phi)

At time t=0 since mass is at it's maximum position thus we have

A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})

Thus the position of mass at the given times is as follows

1) at \frac{\pi}{12} x(t)=5.99inches

2) at \frac{\pi}{8} x(t)=5.9909inches

3) at \frac{\pi}{6} x(t)=5.98397inches

4) at \frac{\pi}{4} x(t)=5.9639inches

5) at \frac{9\pi}{32} x(t)=5.954inches

4 0
2 years ago
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

6 0
2 years ago
Read 2 more answers
An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a
wlad13 [49]
Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
fcp = 66.666... = 66 \: newtons

Answer: Last option, 66 N.
7 0
2 years ago
Read 2 more answers
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80
Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0
2 years ago
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KIM [24]
The answer is:
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2 years ago
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