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2 years ago

3. A supersonic jet flying at 145 m/s experiences uniform acceleration at the rate of 23.1 m/s2 for 20.0 s.

Physics

1 answer:

mihalych1998 [28]2 years ago

7 0

Answer:

A.607m/s | B.1.834Mach (speed of sound)

Explanation:

The plane's speed starts out at 145m/s, meaning after we calculate how many m/s it has accelerated, we must add that to the total.

If it accelerates for 20 seconds at a rate of 23.1m/s^2, this means that it accelerates 23.1m/s/s, or in words it gains 23.1 m/s every second. So, to calculate the gain in speed we simply multiply this by the duration of 20 seconds and get 462 m/s. Now, we add this to 145m/s, and get an end velocity of 607m/s.

For the plane's speed in terms of speed of sound, we divide 607m/s by the speed of sound, or 331m/s, and get ~1.834 Mach(speed of sound)

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A firefighter with a weight of 756 N slides down a vertical pole with an acceleration of 2.96 m/s2, directed downward. What are

Katen [24]

Answer:

a) F = 527.65 N, Force applied is upwards.

b)F = - 527.65 N, where, negative sign depicts Force is applied downwards.

Explanation:

Data provided:

Weight of the firefighter = 756 N

Mass of the firefighter = 756/9.8 = 77.14 Kg

Acceleration, a = 2.96 m/s²

a) In the absence of the pole the firefighter would have been moving down with an acceleration of 9.8 m/s (i.e the acceleration due to the gravity), but due to the presence of the pole the acceleration of the firefighter has been reduced. thus, a force is applied by the pole on the firefighter to reduce the acceleration.

therefore, we have

F = ma(net) = 77.14 × (9.8-2.96) = 527.65 N, Force applied is upwards.

B) According to the Newton's third law, the force will be equal and opposite to the force in the part a)

thus, we have

F = - 527.65 N

3 0

3 years ago

Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

svlad2 [7]

Answer:

Woke done, W = 4156.92 Joules

Explanation:

The work done by the force can be calculated as :

$W=F\times s$

$W=Fs\ cos\theta$

$\theta$ is the angle between force and the displacement

It is assumed to find the work done for the given parameters i.e.

Force, F = 30 N

Distance travelled, s = 160 m

Angle between force and displacement, $\theta=30$

Work done is given by :

$W=Fs\ cos\theta$

$W=30\times 160\ cos(30)$

W = 4156.92 Joules

So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.

5 0

2 years ago

If I bought 59 kids off the black market in a secret door in the stalls and I sell 21 kids then ask for 81 more then 12 off the

larisa [96]

You will be left with 106 kids

<h3>Meaning of word problem</h3>

A word problem can be defined as a mathematical problem that is written in word or written in a sentence format.

In a word problem, the student is expected to decode the sentence into a mathematical expression before solving

In conclusion, You will be left with 106 kids

Learn more about word problems: brainly.com/question/13818690

#SPJ1

5 0

1 year ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago

At what angle are the electronic and the magnetic wave related in an electromagnetic signal?

VikaD [51]

Answer:

90degrees I'm pretty sure

7 0

2 years ago