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Naya [18.7K]
3 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]3 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]3 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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A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.
MaRussiya [10]

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm=\frac{20}{100}=0.20m

1 m=100 cm

Mass of rod,M=190 g=\frac{190}{1000}=0.19kg

Mass of ball,m=19 g=\frac{19}{1000}=0.019 kg

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a  pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

I_{rod-clay}=I_{rod}+I_{clay}

I_{rod-clay}=\frac{1}{3}ML^2+mL^2

Substitute the values then we get

I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2

I_{rod-clay}=3.29\times 10^{-3} kgm^2

Now, the center of mass of the combination of the rod and clay is given by

y=\frac{Md_1+md_2}{M+m}

Substitute d_1=\frac{L}{2}=Distance between pivot and the center of the rod

d_2=L=The distance  between rod and clay

Using the formula

y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}

y=0.1091 m

Time period of the oscillation of the system of the rod and the clay is given by

T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}

g=9.8m/s^2

Using the formula

Time period=2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}

Time-period=0.76 s

Hence, the period =0.76 s

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a 1500.0 kg car is at rest on a level track. The driver gives the engine gas and the car begins to accelerate forward. The car m
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Acceleration of the car is 3.375 m/s² and the force of the car moving forward is 5062.5 N

Explanation:

  • Acceleration is the rate of change of velocity.
  • It is given by the equation, a = change in velocity/time

Here, velocity changes from 0 to 27 m/s and time = 8

⇒ Acceleration = 27 - 0/8 = 27/8 = 3.375 m/s²

  • Force is calculated by the equation, F = Mass × Acceleration
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Here, mass of the car = 1500 kg and a = 3.375 m/s²

⇒ Force = 1500 × 3.375 = 5062.5 N

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