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2 years ago

Find the speed of sound in air at 0 ∘C ( 1100 ft/s ) in mph.

1 answer:

7 0

We are already given the magnitude of the speed which is 1,100 ft/s. All we have to do is convert this to miles per hour. So, we have to know the conversion factor. Note that there are 5,280 ft in 1 mile, and 3,600 s in 1 hr. The solution is as follows:

(1,100 ft/s)(1 mile/5,280 ft)(3,600 s/1 h) = 750 mph

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A FBD of a rocket launching into space should include:

Vladimir [108]

Answer:

Explanation:

the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.

3 0

1 year ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun

AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

$i_1+i_3=i_2$ (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

$i_3=i_2-i_1=0.75A-0.40A=0.35A$

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

$0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}$

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0

2 years ago

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α Formula (1)

where:

τ : It is the moment applied to the body. (Nxm)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :

ωf = ω₀+ α*t Formula (2)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (rad)

Data

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel

We replace data in the formula (2):

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F = tangential force (N)

R = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0

2 years ago

A person standing for a long time gets tired when he does not appear to do any work .why?

Keith_Richards [23]

Explanation:

A person standing still for a long time feels tired because the force of gravity acts on our body and puts stress on our muscles. so our muscles need energy to do work and keep body balanced and help to stand upright.

6 0

2 years ago

pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

solution:

given:

p_o=1.07*10^5 N/m^2

ρ_infinity=1.23 kg/m^2

p_infinity=1.01*10^5 N/m^2

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0

2 years ago