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4 years ago

12

Why do we have 365 days in a year?

1 answer:

uranmaximum [27]4 years ago

4 0

We have 365 days in a year because that is the approximate time of the world revolving around the sun once. It takes 365.25 days to go around once, but the calendar is 365, so we add leap years to add a couple more days.

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Which chemical reaction absorbs energy?

andre [41]

Photosynthesis. Photosynthesis is the process the plant uses to absorbs light to make food

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3 years ago

Read 2 more answers

What type on msd causes inflammation of the tendons that correct bones to muscles

svp [43]

I believe it would be Tendonitis

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4 years ago

Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the

viva [34]

Answer:

V=19.08 m/s

Explanation:

Airplane gross weight w=7780 N

Airplane wing area S=16.6 m²

Air density p=1.2250 kg/m³

Maximum lift coefficient CL=2.1

To find

Stalling Speed

Solution

The equation to find stalling speed is given below

$W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\ V_{Stalling-speed}=19.08 m/s$

5 0

3 years ago

A copper sphere 10 mm in diameter is dropped into a 1-m-deep drum of asphalt. The asphalt has a density of 1150 kg/m3 and a visc

kvasek [131]

Answer:

t = 1964636.542 sec

Explanation:

Given data:

sphere diameter is 10 mm

Density is 1150 kg/m^3

viscosity 105 N s/m^2

We knwo that time taken by sphere can be calculated by following procedure

$\tau = \mu \frac{du}{dy}$

$\frac{F}{A} = \mu \frac{du}{r}$

$\frac{\rho_C -\rho_{asphalt} gv}{2 \pi rL} = 10^5 \frac{du}{r}$

Solving for du

$du = \frac{ (8933 - 1150) 9.81 \frac{4}{3} \pi (10\times 10^{-3})^3}{2\pi \times 1\times 10^5}$

$du = u = 5.09\times 10^{-7}$

$u = \frac{1}{t}$

$t = \frac{1}{5.09\times 10^{-7}} = 1964636.542 sec$

6 0

4 years ago

A particle leaves the origin with an initial velocity v = (3.00 i hat) m/s and a constant acceleration a = (-3.00 i hat - 1.400

Brilliant_brown [7]

Answer:

a) —0.5 j m/s

b) 4.5 i + 2.25 j m

Explanation:

<u>Givens:</u>

v_0 =3.00 i m/s

a= (-3 i — 1.400 j ) m/s^2

The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :

<em>v_x = v_0 + at = 0 </em>

(3.00 i m/s) + (-3 i m/s^2)t=0

t = (3 m/s)/-3 i m/s^2

t = -1 s

Therefore the particle reaches the maximum x-coordinate at time t = 1 s.

Part a The velocity-of course- is all in the y-direction,therefore:

v_y =v_0+ at

We have that v_0 = 0 in the y-direction.

v_y = (-0.5 j m/s^2)(1 s)

= —0.5 j m/s

Part b: While the position of the particle at t = 1 s is given by:

r=r_0+v_0*t+1/2*a*t^2

Where r_0 = 0 since the particle started from the origin.

Its position at t = 1 s is then given by :

r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2

=4.5 i + 2.25 j m

8 0

3 years ago