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mash [69]
2 years ago
8

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

the depth in the rapids is 0.85m.Find the speed of water flowing in the rapids, given that its speed before the rapids is 1.2 m/s. Assume the river has a rectangular cross section.
Physics
1 answer:
Alisiya [41]2 years ago
6 0

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

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VikaD [51]

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

V = IR \rightarrow R = \frac{V}{I}

Our values are

I = 500mA = 0.5A

V = 37V

Replacing,

R = \frac{V}{I}

R = \frac{37}{0.5}

R = 74 \Omega

Therefore the smallest resistance you can measure is 74 \Omega

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3 years ago
Which statement is true for objects in dynamic equilibrium?
lakkis [162]

Answer:

The correct answer to the question is objects have zero acceleration.

Explanation:

Before answering the question, first we have to understand dynamic equilibrium .

A body moving with uniform velocity is said to be in dynamic equilibrium if the net external forces acting on the body is zero. Hence, the body is under balanced forces.

If the external forces acting on a body is not balanced, then the body will accelerate which will destroy its equilibrium condition. Hence, the necessary and sufficient condition for a body to be in dynamic equilibrium is that the forces are balanced.

When a body is in dynamic equilibrium, the body moves with uniform velocity along a straight line unless and until it is compelled by some external unbalanced forces.

Hence, the rate of change of velocity or acceleration of the body will be zero.

5 0
3 years ago
Read 2 more answers
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up
Westkost [7]

Answer:

24 N

Explanation:

m = mass of the cube = 6.0 kg

Consider the three cubes together as one.

M = mass of the three cubes together = 3 m = 3 (6.0) = 18 kg

a = acceleration of the combination = 2 ms⁻²

F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

4 0
3 years ago
If a specimen was being viewed using a 20x objective lens and 10x ocular lens, what would be the total magnification
Paraphin [41]

Answer:

As Per Given Information

20x objective lens was used by specimen

10x ocular lens was also used by him.

we have to find the total magnification.

For calculating the total magnification we 'll simply do multiplication

Total Magnification = 20x × 10x

Total Magnification = 200x

So , the total magnification will be 200x .

6 0
2 years ago
A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
3 years ago
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