Which increases the rate of soil formation?

Find the volume of a box that is 122 cm long, 51 cm wide and 13 cm high.<br> V= ________cm³

DochEvi [55]

Answer:

80,886 cm^3

Explanation:

v= lwh (also known as length × width × height)

v = 122cm × 51cm × 13cm

v = 80,886 cm^3

6 0

4 years ago

A convex mirror with a focal length 20 cm creates a virtual image 10cm away from the mirror. What is the position of the object?

nikdorinn [45]

Answer:

$20$ cm

Explanation:

$d_{o}$ = distance of the object = ?

$f$ = focal length of the convex mirror = - 20 cm

$d_{i}$ = distance of the image = - 10 cm

Using the lens equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{d_{o}} + \frac{- 1}{10} = \frac{- 1}{20}$

$d_{o} = 20$ cm

6 0

3 years ago

A musician in a group plays a wrong note. would this note disrupt the harmony of the rhythm of the song being played? explain yo

Eddi Din [679]

If a musician plays a wrong note, he will disrupt the harmony of rhythm of the song. This is because, he will be able to produce a sound of different wavelength which will cause a possible destructive interference with the waves being produced by his groupmates.

8 0

3 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch

Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If ' $W$ ' be the work done on an object to change its kinetic energy from an initial value ' $K_{i}$ ' to the final value ' $K_{f}$ ', then mathematically,

$W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)$

where ' $m$ ' is the mass of the object and ' $v_{i}$ ' and ' $v_{f}$ ' be the initial and final velocity of the object respectively. If ' $F_{net}$ ' be the net force applied on the car, as per given problem, and ' $s$ ' is the displacement occurs then we can write,

$W = F_{net}~.~s.......................................................(II)$

Given, $m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m$ .

Equating equations (I) and (II),

$&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N$

6 0

3 years ago