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romanna [79]
1 year ago
13

While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is

not completed before crossing the road or reference line. This would most likely occur in turn:_______
Physics
1 answer:
erica [24]1 year ago
5 0

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.  

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

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A projectile of mass m is fired horizontally with an initial speed of v0​ from a height of h above a flat, desert surface. Negle
Grace [21]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

5 0
2 years ago
When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180
Bess [88]

Answer:

b

Explanation:

i took the test

6 0
1 year ago
Read 2 more answers
The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's
Mice21 [21]

Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

Frequency of the piano, f = 440 Hz

Length of the piano, L = 38.9 cm = 0.389 m

Tension in the spring, T = 667 N

The frequency in the spring is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{L} is the linear mass density

On rearranging, we get the value of m as follows :

m=\dfrac{T}{4Lf^2}

m=\dfrac{667}{4\times 0.389\times (440)^2}

m = 0.0022 kg

or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.

3 0
2 years ago
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Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
2 years ago
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A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
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