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3 years ago

What is Average rainfall in the desert?

Physics

2 answers:

Vadim26 [7]3 years ago

7 0

Answer:

less than 10 inches, or 25 centimeters, of precipitation a year.

Explanation:

PS. the precipitation recieved in the desert can be in the form of either rain or snow.

pav-90 [236]3 years ago

3 0

It is less than 10 Inches from what I have done my research project on

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A box with mass m = 8 kg is pushed x = 10 m across a level floor by a constant applied force F P = 16.27 N. The coefficient of k

vitfil [10]

Answer:

Final speed of the box after it has moved to x = 10 is given as

v = 3.35 m/s

Explanation:

As we know by work energy theorem that work done by all the forces is equal to the change in its kinetic energy

Here work done by external force + work done by friction = change in kinetic energy of the box

so we have

$W_{ex} + W_{fric} = \frac{1}{2}mv^2$

$F x - \mu mg x = \frac{1]{2}mv^2$

$16.27 (10) - (0.15)(8)(9.8) (10) = \frac{1}{2}(8) v^2$

$162.7 - 117.6 = 4 v^2$

$v = 3.35 m/s$

4 0

3 years ago

The momentum of an object is determined to be 7.2 x 10-3 cm kg x m/s. Express this as provided or use any equivalent unit. How i

Leno4ka [110]

Complete Question:

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kg = 1000 g).

Answer:

7.2 gm/s.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

Given the following data;

Momentum = 7.2 * 10^-3 kgm/s

1 kg = 1000 g

Substituting the unit in kilograms with grams, we have;

Momentum = 7.2 * 10^-3 * 1000 gm/s

<em>Momentum = 7.2 gm/s. </em>

7 0

3 years ago

Compare an Earth year to a cosmic year

sweet-ann [11.9K]

Compared with an Earth year, a galactic year represents time on a grand scale but its not a consistent measurement across the galaxy

4 0

3 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

What must a system owner do, once the threshold leak rate has been exceeded on any low-pressure system using an ozone-depleting

soldier1979 [14.2K]

Answer:

A procedure according to the norms.

Explanation:

If possible, proceed to fix the leak in no more than 30 days from the moment it was discovered.

Otherwise, during the first 30 days develop a planification to backfit the leak or, if needed, retire the appliance. This should be executed within one year.

7 0

3 years ago