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2 years ago

15

A car is moving along a straight line op is shown below it moves from O to P in 8sec and return to p to q in 6sec. What is the a

verage velocity and the speed of the car

1 answer:

Pavel [41]2 years ago

4 0

Answer:

The answer would be A

Explanation:

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A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds

melisa1 [442]

Answer:

C

Explanation:

65-35 = 30

30 divided by 5 = 6

The answer will be 6 m/s

6 0

2 years ago

Two streams merge to form a river. One stream has a width of 8.4 m, depth of 3.5 m, and current speed of 2.2 m/s. The other stre

hoa [83]

Answer:

Explanation:

We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .

rate of volume flow of one stream

= cross sectional area x velocity

= 8.4 x 3.5 x 2.2 = 64.68 m³ /s

rate of volume flow of other stream

= 6.6 x 3.6 x 2.7

= 64.15 m³ /s

rate of volume flow of rive , if d be its depth

= 11.2 x d x 2.8

= 31.36 d

volume flow of river = Total of volume flow rate of two streams

31.36 d = 64.15 + 64.68

31.36 d = 128.83

d = 4.10 m /s .

6 0

3 years ago

Renewable energy sources are ____________________?

Nina [5.8K]

Answer:

Eco-friendly

Explanation:

6 0

3 years ago

If a car travels 600m west in 25 seconds, what is its velocity?

Elena L [17]

$v = \frac{ \:the covered distance}{time} \\$

$v = \frac{600}{25} = \frac{6 \times 100}{25} = 6 \times 4 = 24 \: \\$

$v = 24 \: \frac{m}{s} \\$

_________________________________

If west means the west of the axis x the velocity equal :

$- 24 \: \frac{m}{s} \\$

4 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago