What is the net force acting on this box?

A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief

fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

$x_1 = -120cos70^o = -41.04 km$

$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

3 0

3 years ago

Two 4.3546 cm x 4.3546 cm square aluminum electrodes, spaced 0.6408 mm apart are connected to a 73.68 V battery. What is the cap

Helen [10]

Answer:

Capacitance, C = 26.1 picofarad

Explanation:

It is given that,

Side of square, x = 4.3546 cm = 0.043546 m

Distance between electrodes, d = 0.6408 mm = 0.0006408 m

Voltage, V = 73.68 V

Capacitance of parallel plates is given by :

$C=\dfrac{\epsilon_oA}{d}$

$C=\dfrac{8.85\times 10^{-12}\times (0.043546)^2}{0.0006408}$

$C=2.61\times 10^{-11}\ F$

or

C = 26.1 picofarad

So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.

3 0

3 years ago

A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released

statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

mass of the block, m = 2.4 kg
height of the block, h = 5 m
compression of the spring, x = 25 cm = 0.25 m

The spring constant is calculated as follows;

$F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m$

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

$\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2 = mgh - \frac{1}{2} kx^2\\\\mv^2 = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s$

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0

3 years ago

The students in the diagram wanted to calculate the speed of sound. They

topjm [15]

Isolate the variable by dividing each side by factors that don't contain the variable.

m

=

32.021

¯

3

s

8 0

3 years ago

3500kg of water are poured into a water tank and fill a volume equal to 3.5m^3. a. What is the density of the water?

Vaselesa [24]

Answer:

a )

Density = mass / volume

= 3500 / 3.5

= 1000 kg m⁻³

b) gauge pressure = h d g

h is height of water column

d is density

g is acceleration due to gravity

Put the values in the equation above

Gauge pressure = .15 x 1000 x 9.8

= 1470 Nm⁻²

Atmospheric pressure = 100000 Nm⁻²

Total pressure

= 101470 N m⁻²

c ) Volume of water displaced = volume of object

= 55 x 10⁻⁶ m³

Buoyant Force = weight of displaced water

= 55 x 10⁻⁶ x 1000 x 9.8

= .539 N.

8 0

3 years ago