what is the mass of of a wooden block that has a length of 4cm and a width of 5cm and a height of 10cm

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

il63 [147K]

<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

6 0

3 years ago

Which of the following does not illustrate a complete circuit?

Yakvenalex [24]

Answer:

I think A

Explanation:

because it dosn't have enough tools

4 0

2 years ago

The plates of a spherical capacitor have radii 6.25 cm and 15.0 crn. The space between the two spheres is filled with a material

aleksklad [387]

Answer:

Capacitance is 0.572×10⁻¹⁰ Farad

Explanation:

Radius = R₁ = 6.25 cm = 6.25×10⁻² m

Radius = R₂ = 15 cm = 15×10⁻² m

Dielectric constant = k = 4.8

Electric constant = ε₀ = 8.854×10⁻¹² F/m

ε/ε₀=k

ε=kε₀

$Capacitance\ (C)=\frac{4\pi k\epsilon_0 R_1\times R_2}{R_2-R_1}\\\Rightarrow C=\frac{4\pi 4.8\times 8.854\times 10^{-12}\times 15\times 10^{-2}\times 6.25\times 10^{-2}}{15\times 10^{-2}-6.25\times 10^{-2}}\\\Rightarrow C=0.572\times 10^{-10}\ Farad$

∴ Capacitance is 0.572×10⁻¹⁰ Farad

3 0

3 years ago

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

3 years ago

If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo

Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0

2 years ago