Ask question

Ask question

All categories

3 years ago

13

a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th

e target fire should the load be released?

1 answer:

Andreyy893 years ago

3 0

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

You might be interested in

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small

Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0

2 years ago

You have a length of tubing that is closed at one end. You cut the tubing into two pieces of unequal length, giving you a tube o

umka21 [38]

Answer:

The funda mental frequency of the original tube is 182Hz.

Explanation:

See the attachment for the calculation steps.

In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.

For closed tubes

f = nv/4L (n = 1, 3, 5,...n)

f = nv/2L (n = 1, 2, 3,...n)

The details of calculation can be found below in the attachment.

4 0

3 years ago

In what two ways can we increase the potential difference?

vredina [299]

Yes I hope this helps

4 0

3 years ago

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s

andre [41]

Answer:

$d=1.84\ mm$

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

$\displaystyle C=\frac{\epsilon_o A}{d}$

where

$\epsilon_o=8.85\cdot 10^{-12}\ F/m$

A = area of the plates = $\pi r^2$

d = separation of the plates

$\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}$

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

$\displaystyle X_c=\frac{1}{wC}$

where w is the angular frequency

$w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s$

Solving for C

$\displaystyle C=\frac{1}{wX_c}$

The reactance can be found knowing the total impedance of the circuit:

$Z^2=R^2+X_c^2$

Where R is the resistance, $R=15 K\Omega=15000\Omega$ . Solving for Xc

$X_c^2=Z^2-R^2$

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

$\displaystyle Z=\frac{V}{I}$

The rms current is the peak current Ip divided by $\sqrt{2}$ , thus

$\displaystyle Z=\frac{\sqrt{2}V}{I_p}$

$I_p=0.65\ mA/1000=0.00065\ A$

Now collect formulas

$\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2$

Or, equivalently

$\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}$

$\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}$

$X_c=36176.34\ \Omega$

The capacitance is now

$\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F$

The radius of the plates is

$r=18\ cm/2=9 \ cm = 0.09 \ m$

The separation between the plates is

$\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}$

$d=0.00184\ m$

$\boxed{d=1.84\ mm}$

8 0

2 years ago

A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

Lina20 [59]

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop $A=\pi r^2$

$A=3.14\times 0.045^2=0.00635m^2$

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field $dB=250-350=100mT$

Emf induced is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V$

Magnitude of induced emf is equal to 0.00635 V

7 0

3 years ago