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Sholpan
1 month ago
13

An airplane cruising at a constant velocity and altitude. Which of the following diagrams best represents the four forces of the

airplane?
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
7 0

Answer:

Please include the diagrams and repost them.

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A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Sav [3153]
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
14 days ago
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i
Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

a=\frac{v^2}{R}

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = 9.81\ m/s^2

v = ?

R = 13.2 m

Por lo tanto,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

v = 11.4 m/s

8 0
1 month ago
If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.
ValentinkaMS [3465]
V = I * R, where V signifies voltage, I represents current, and R is resistance. According to Ohm's law, to determine the current through the wire, resistance is required. In theory, if the wire had zero resistance, it would lead to infinite current, which is not feasible. This negligible resistance could refer to the internal resistance of the battery rather than the wire itself. 
7 0
14 days ago
Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
ValentinkaMS [3465]

Answer:

20 cm

Explanation:

The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².

Thus, solving for r gives us r = kq₁q₂/U

which leads to x - 2 = kq₁q₂/U

Then, rearranging gives x = 0.02 + kq₁q₂/U m

So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm

7 0
16 days ago
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
1 month ago
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