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3 years ago

An airplane is flying in air with a density of 1.29 kg/m3. A pressure gauge measures

Physics

1 answer:

jeka57 [31]3 years ago

5 0

Answer:

341 m/s

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

The velocity of the air at the nose is 0 m/s, so:

P₁ = P₂ + ½ ρ v₂²

ΔP = ½ ρ v₂²

Plugging in values:

75000 Pa = ½ (1.29 kg/m³) v²

v = 341 m/s

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mojhsa [17]

Answer:

Explanation:

There are 3 main forces at work here, gravity, normal and friction. The gravity pulls the car straight down and is what keeps the car on the ground. Normal force is straight up from the points where the car is touching, so since the wheels are the only parts of the car touching the street, this is where all the normal force is. Friction force opposes any and all motion, the car wants to slide down the hill and would slide down the hill if there was no friction, so the friction force is in the opposite direction of the cars intended motion.

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3 years ago

The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant

KIM [24]

Answer: 50 km, 0, 27.78 m/s

Explanation:

Given

Circumference of the track is $12.5\ km$

Speed of car is $100\ km/h$

Car drives for $30\ \text{minute}\ or\ 0.5\ hr$

(a)Distance traveled is

$\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km$

(b)displacement of the car

It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.

Therefore, the displacement of the car is zero.

(c)To convert kmph to m/s, multiply the entity by $\frac{5}{18}$

$\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s$

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3 years ago

Two identical parallel plate capacitors A and B connected to a battery of V volts w/ the switch S closed. The switch is now open

Basile [38]

The electrostatic energy stored in a capacitor with capacitance $C_0$ , with a voltage difference V applied to it, and without dielectric, is given by
$U_0 = \frac{1}{2} C_0 V^2$
Now let's assume we fill the space between the two plates of the capacitor with a dielectric with constant k. The new capacitance of the capacitor is
$C_k = k C_0$
So, the energy stored now is
$U_k = \frac{1}{2}C_k V^2= \frac{1}{2}kC_0 V^2$

Therefore, the ratio between the energies stored in the capacitor before and after the introduction of the dielectric is
$\frac{U_k}{U_0}= \frac{ \frac{1}{2}kC_0 V^2 }{ \frac{1}{2} C_0 V^2}= k$

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3 years ago

an NHL hockey player (m=90 kg) stands motionless on the ice . the 10 kg stanley cup is thrown to the player at 4m/s. after he ca

spayn [35]

Answer:v=0.4 m/s

Explanation:

Mass hockey player $m_1=90 kg$