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BARSIC [14]
4 years ago
9

A boat travels between two riverside cities, A and B, located 21 miles apart. The river flows in the direction from A to B at 2

mph. The boat takes a total of 10 hours to make a round trip from A to B and back to A again. How long does it take the boat to make the one-way trip from A to B if the boats speed remailer constant relative to the speed of the river in each direction
Physics
1 answer:
klasskru [66]4 years ago
4 0

Answer:

The time taken by the boat to make the one-way trip from A to B = 3 hrs

Explanation:

Distance between A & B = 21 mile

Velocity of river ( v ) = 2 \frac{mi}{hr}

Total time taken ( T )= 10 hours

T = \frac{D}{u - v}  + \frac{D}{u +v}

10 = \frac{21}{u - 2} +  \frac{21}{u + 2}

10 = \frac{42 u}{u^{2}-4 }

10 u^{2} - 42 u -40 = 0

u^{2} - 4.2 u - 4 = 0

By solving above equation we get

u = 5 \frac{mi}{hr}

The time taken by the boat to make the one-way trip from A to B

T = \frac{D}{u + v}

T = \frac{21}{5 + 2}

T = 3 hours

This is the time taken by the boat to make the one-way trip from A to B.

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Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of
Alchen [17]

Answer:

\theta=10.20^{\circ}  

\Delta l=0.10 ft    

Explanation:

First of all, we analyze the system of blocks before starting to move.

\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

16sin(\theta)-2.91cos(\theta)=0  

tan(\theta)=0.18  

\theta=arctan(0.18)  

\theta=10.20^{\circ}  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

P_{A}sin(\theta)-F_{fA}-F_{spring}=0

Where:

F_{spring} = k\Delta l=2.1\Delta l

P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

\Delta l=0.10 ft    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

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4 years ago
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A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the c
Goryan [66]

Answer:

<em>The velocity of the truck is 3.33 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If some collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses:

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There are two objects: The m1=4000 Kg car and the m2=6000 Kg truck. The car was moving initially at v1=4 m/s and the truck was at rest v2=0. After the collision, the car moves at v1'=-1 m/s. We need to find the velocity of the truck v2'. Solving for v2':

\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}

Substituting:

\displaystyle v'_2=\frac{4000*4+6000*0-4000(-1)}{6000}

\displaystyle v'_2=\frac{16000+4000}{6000}

\displaystyle v'_2=3.33

The velocity of the truck is 3.33 m/s

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Explanation:

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Specific heat is an intensive property and does not depend on the amount of matter that is present within a substance.

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