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3 years ago

13

Because scientists cannot actually precisely measure a process like soil erosion over the entire planet, they must _____.

1 answer:

Pani-rosa [81]3 years ago

4 0

Your answer is C. I hope this helps you.

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A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.

g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle $=34^{\circ}$

Range=240 m

We know that Range $=\frac{u^2sin2\theta }{g}$

$240=\frac{u^2sin(68)}{9.81}$

u=50.39 m/s

(b)maximum height of projectile is given by

$H=\frac{u^2sin^2\theta }{2g}$

$H=\frac{50.39^2(sin34)^2}{2\times 9.81}$

H=40.46 m

3 0

3 years ago

Read 2 more answers

A passenger on a stopped bus notices that rain is falling vertically just outside the window. When the bus moves with constant v

nekit [7.7K]

Answer:

1)0.325

2) $6.17\ \rm m/s$

Explanation:

<u>Given:</u>

The angle that falling raindrops make with the vertical= $18^\circ$

Let $V_R$ be the velocity of the raindrops and $V_B$ be the velocity of the bus.

1)

$\dfrac{V_R}{V_B}=\tan 18^\circ\\\dfrac{V_R}{V_B}=0.315\\$

2)Speed of the raindrops $=0.315\times 19$

$=6.17\ \rm m/s$

5 0

2 years ago

Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you

gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

$\mu_{0} = 4\pi * 10^{-7}$

The inductance of a solenoid is given by:

$L = \frac{\mu_{0}N^{2} A }{l}$

$500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns$

8 0

2 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

2 years ago

A CD has to rotate under the readout-laser with a constant linear velocity of 1.25 m/s. If the laser is at a position 3.7 cm fro

Savatey [412]

Answer:N=322.53 rpm

Explanation:

Given

Linear velocity (v)=1.25 m/s

Position from center is 3.7 cm

we know

$v=\omega \times r$

$1.25\times 100=\omega \times 3.7$

$\omega =\frac{125}{3.7}=33.78$

and $\frac{2\pi N}{60}=\omega$

$N=\frac{\omega \times 60}{2\pi }$

$N=\frac{33.78\times 60}{2\pi }$

N=322.53 rpm

8 0

3 years ago