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Airida [17]
3 years ago
9

A boat travels upstream for 12 miles in 3 hours and returns in 2 hours traveling downstream in a river. What is the rate of the

boat in still water and the rate of the current?
Physics
1 answer:
kaheart [24]3 years ago
7 0

Answer:

Rate of boat in still water = 5mile/hour

Rate of the current= 1 mile/hr

Explanation:

The boat is moving upstream at the rate= 12miles/3hours= 4mph

The boat is moving downstream at the rate = 12miles/2hours = 6mph

Let y be the speed of boat

And C the current of water

Y + C = 6 ...eq1

Y - C = 4 ...eq2

Adding eq1 and eq2

2Y = 10

Y= 10/2 = 5

Y= 5mph

Put Y = 5 in eq2

5 + C = 6

C = 6 - 5

C= 1 mph

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lord [1]

Answer:

One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.

4 0
3 years ago
Evaluate u+xy where U=3 X=4 Y=7
Reptile [31]

Answer:

31

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

31

6 0
2 years ago
What is the primary type of battery we use today to store energy?
Ahat [919]

Lithium-Ion batteries are commonly used in portable electronics and electric vehicles. These rechargeable batteries have two electrodes: one that's positively charged and contains lithium and another negative one that's typically made of graphite.

4 0
2 years ago
Read 2 more answers
h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round t
laila [671]

Answer:

2 seconds

Explanation:

The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.

h(t)=-16t^{2}+64t+112

Differentiate with respect to t on both the sides, we get

\frac{dh}{dt}=-32t+64

For maxima and minima, put the value of dh / dt is equal to zero. we get

- 32 t + 64 = 0

t = 2 second

Thus, the arrow reaches at maximum height after 2 seconds.

8 0
3 years ago
6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750
zysi [14]

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

v_{f}^{2}=v_{i}^{2} +2*a*x

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]

7 0
2 years ago
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