Ask question

Ask question

All categories

3 years ago

9

When we study the earth and its people, we are studying ___________________. a) desalination b) geography c) irrigation d) absol

ute location

2 answers:

Fofino [41]3 years ago

7 0

Geography Earth studying

lbvjy [14]3 years ago

7 0

The answer is b geography

You might be interested in

A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1600 kPa. The working flui

anzhelika [568]

Answer:

a) 6498.84 kW

b) 0.51

c) 0.379

Explanation:

See the attached picture below for the solution

7 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si

JulijaS [17]

Answer:

0.75

Explanation:

$refractive \: index \: = \frac{3.0 \times {10}^{2} }{2.0 \times {10}^{2} }$

= 1.5

$refractive \: index = \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }$

$1.5 = \ \frac{ \sin(i) }{ \sin(30) }$

1.5 × ½ = sin(i)

$\sin(i) = 0.75$

5 0

3 years ago

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = $\frac{1}{2} kx^2$ and the kinetic energy gained by the box is given by KE = $\frac{1}{2} mv^2$

Since work done by the spring = kinetic energy gained by the box we have

$\frac{1}{2} mv^2$ = $\frac{1}{2} kx^2$ therefore we have v = $\sqrt{\frac{kx^2}{m} }$ = $x\sqrt{\frac{k}{m} }$ = $0.15\sqrt{\frac{300}{0.2} }$ = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = $\frac{v^2}{2g}$ = $\frac{5.81^{2} }{2*9.81}$ = 1.72 m

6 0

3 years ago

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference

strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

$n_{air}=1$

$\theta_{liquid} = 19.38\°$

$\theta_{air}35.09\°$

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

$n_1sin\theta_1 = n_2sin\theta_2$

$n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}$

$n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}$

Replacing the values we have:

$n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}$

$n_liquid = 1.7323$

Therefore the refractive index for the liquid is 1.7323

6 0

3 years ago