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Lostsunrise [7]

3 years ago

7

A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full

scale airplane in air. The velocity in the wind tunnel is then

1 answer:

trasher [3.6K]3 years ago

7 0

Answer:

of the velocity of a full size plane in the air

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A block moving to the right on a level surface with friction is pulled by an increasing horizontal force also directed to the ri

KengaRu [80]

Answer:

e) True Yes both are constant

Explanation:

Let us propose the solution of the problem before reviewing the statements, we use Newton's second law

F - fr = m a

N- W = 0

N = mg

The equation for the force of friction is

fr = μ N

F - μ mg = m a

F = m (a- μ g)

Now let's review the claims

.a) False. Normal force and friction are constants.

.b) False. Both are constant.

.c) False. Both are constant.

d.) False

e) True Yes both are constant

6 0

3 years ago

An object moves along the x-axis according to the equation x = 3.00t2 – 2.00t + 3.00,

Ray Of Light [21]

Explanation:

x = 3.00 $t^{2}$ – 2.00t + 3.00,

Distance of object at 2 second,

x (t=2) = 3(4) - 2(2) +3

x (t=2) = 12-4 +3

x (t=2) = 11 m

Distance of object at 3 second,

x (t=3) = 3(9) - 2(3) +3

x (t=2) = 27 - 6 + 3

x (t=2) = 24 m

a) the average speed between t = 2.00 s and t = 3.00 s,

Average speed = $\frac{Total distance}{ Total time}$

Average speed = $\frac{x (t=2) + x (t=3)}{3}$

Average speed = $\frac{24+11}{3}$

Average speed = $\frac{35}{3}$

Average speed = 11.66 $\frac{m}{s}$

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = $\frac{dx}{dt}$

Instantaneous speed(v) = 6t - 2 $\left \{ {{t=2} \atop {t=3}} \right.$

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 $\frac{m}{s}$

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = $\frac{average velocity}{time}$

average acceleration = $\frac{11.66}{3-2}$

average acceleration = 11.66 $\frac{m}{s^{2} }$

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = $\frac{dv}{dt}$

instantaneous acceleration =6

instantaneous acceleration = 6 $\frac{m}{s^{2} }$

e) for x =0

0 = 3.00 $t^{2}$ – 2.00t + 3.00

a = 3, b=-2, c=3

t= $\frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

t= $\frac{2 \pm \sqrt{4 - 36} }{6}$

t= $\frac{2 \pm \sqrt{-32} }{6}$

general solution of this equation gives imaginary value. Hence, the given object is not at rest.

7 0

3 years ago

Read 2 more answers

Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT include

Kaylis [27]

The electric filed in V/m is 1.58 * 10^6 V/m

<h3>What is the electric field?</h3>

We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.

Thus;

E = V/d

E = electric field

V = voltage

d = distance of separation

E = 24 * 10^3 V/1.52 * 10^-2 m

E = 1.58 * 10^6 V/m

Learn more about electric field :brainly.com/question/8971780

#SPJ1

7 0

2 years ago

A rocket burns fuel at a rate of 199 kg/s and exhausts the gas at a relative speed of 9 km/s. Find the thrust of the rocket. Ans

andreyandreev [35.5K]

Answer:

1.791 MN

Explanation:

Thrust of the rocket can be found using the relation

T = v.dm/dt, where

T = thrust off the rocket

v = speed of the rocket, 9 km/s = 9000 m/s

dm/dt = rate at which fuel burns, 199 kg/s

Substituting the values into the formula, we have

T = 9000 * 199

T = 1791000 N

T = 1.791*10^6 N

Since 1 MN = 10^6, thus

T = 1.791 MN

7 0

3 years ago

Which statement is true regarding the waves shown?

never [62]

(A) is the answer

Explanation:

hope it help

5 0

3 years ago