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3 years ago

7

Norway, Sweden, and Finland (Scandinavia) have what kind of climates??

2 answers:

boyakko [2]3 years ago

8 0

Answer:

baltic + mild

Explanation:

Romashka [77]3 years ago

3 0

They have a mild climate.

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In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

7 0

2 years ago

If you walk 800m north, 600m east, and 200 meters south at a speed of 1m/s. What is your velocity?

ryzh [129]

Answer:

0.375 m/s north & 0.375 m/s east

Explanation:

8 0

3 years ago

Convert time from 12-hour to 24-hour clock.

Kamila [148]

22:54 is the answer you are looking for

6 0

2 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

2 years ago

A bird is about 6.26.2 in. long, with a thin, dark bill and a wide, white wing stripe. If the bird can fly 9292 mi with the w

Trava [24]

Answer:

$209$ mph

Explanation:

$V$ = Speed of bird in still air

$v$ = Speed of wind = 44 mph

Consider the motion of the bird with the wind

$D_{1}$ = distance traveled with the wind = 9292 mi

$t_{1}$ = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

$t_{1} = \frac{D_{1}}{V + v}$

$t_{1} = \frac{9292}{V + 44}$ eq-1

Consider the motion of the bird with the wind

$D_{2}$ = distance traveled against the wind = 6060 mi

$t_{2}$ = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

$t_{2} = \frac{D_{2}}{V + v}$

$t_{2} = \frac{6060}{V - 44}$ eq-2

As per the question,

Time taken with the wind = Time taken against the wind

$t_{1} = t_{2}$

$\frac{9292}{V + 44} = \frac{6060}{V - 44}$

$(9292) (V - 44) = (6060) (V + 44)$

$9292V - 408848 = 6060V + 266640$

$3232V = 675488$

$V = 209$ mph

5 0

2 years ago