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love history [14]
3 years ago
8

27. The pressure at A is 9.5 atm and the water velocity is 10 m/s. What is the water velocity at point C

Physics
1 answer:
pogonyaev3 years ago
6 0

Answer:

10 m/s

Explanation:

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A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rota
wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

Length of beam = 4.0 m

Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

Point at which person stands = 1.50 m from wall

Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

T×sin(53) = 350×1.5 + 200×2

T = 850/sin(53)  = ‭1,064.315 N.

4 0
2 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk
ahrayia [7]

Answer:

C) 1.0 m

Explanation:

The component of the velocity parallel to the sidewalk is:

vₓ = v cos θ

vₓ = 0.1 m/s cos 45°

vₓ = 0.0707 m/s

The distance traveled after 14 seconds is:

d = vₓ t

d = (0.0707 m/s) (14 s)

d = 0.99 m

Closest answer is C) 1.0 m.

6 0
3 years ago
Read 2 more answers
In 1-2 sentences explain how you would determine the constant k for a spring.
Dmitrij [34]

To perform an experiment to determine the force constant of a spring, you will need a stand with a boss and clamp, a spiral spring, a meter rule and different weights.

The setup is arranged as shown in the image attached. The natural length of the spring is first recorded. Different weights are added to the spring one after the other and the extension is recorded.

The weight is now plotted on the vertical axis and the extension is plotted on the horizontal axis. The slope of the graph is the force constant of the spring.

Learn more: brainly.com/question/10991960

3 0
3 years ago
n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad
Oksi-84 [34.3K]

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

5 0
3 years ago
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