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stealth61 [152]

3 years ago

9

Deodorants use materials that sublime easily. Why does the rate of sublimation of these materials increase after they are applie

d to the human body?

2 answers:

umka21 [38]3 years ago

6 0

Because some people bodies or different, and people sweat a lot depends on what they are doing and how active they are.

Gre4nikov [31]3 years ago

3 0

Explanation:

Sublimation is a process in which a solid directly changes into liquid phase without undergoing liquid phase or vice versa.

Therefore, when we apply a deodorant then its particles come in contact with the surroundings or atmosphere.

As a result, they diffuse at a faster rate and hence, they spread quickly when applied to the human body.

You might be interested in

A rock drops from a height of 60 m. how long does it take for it to hit the ground

Svet_ta [14]

Here, Apply 2nd equation of the Kinematics :
S = ut + 1/2 at²
Here, s = 60 m
u = 0 [ free fall ]
a = 9.8 m/s² [ constant value for Earth system ]

Substitute their values,
60 = 0*t + 1/2 * 9.8 * t²
120 = 9.8t²
t² = 120 / 9.8
t = √12.24
t = 3.50 s

In short, Your Answer would be 3.50 seconds

Hope this helps!

7 0

3 years ago

A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g 5 9.80 m/s2. Find (a) the period of the pend

choli [55]

Answer:

(a) 1.5 second

(b) 0.56 m

Explanation:

Pendulum makes 120 oscillations in 3 min that means in 180 seconds

time taken by the pendulum to complete one oscillation is called time period.

(a) So, the time period is 180 / 120 = 1.5 second

T = 1.5 second

Thus, the time period of the pendulum is 1.5 second.

(b) g = 9.8 m/s^2

The formula for the time period is given by

$T =2\pi \sqrt{\frac{L}{g}}$

Where, L be the length of pendulum

$1.5 =2\times 3.14 \sqrt{\frac{L}{9.8}}$

$0.057= {\frac{L}{9.8}}$

L = 0.56 m

Thus, the length of the pendulum is 0.56 m .

6 0

3 years ago

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

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

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

B)

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$

We can verify that at the same time, the vertical position of object 1 is the same:

$y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m$

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

$v_{1x}=25 m/s$

So its horizontal position at t = 2.0 s is

$x_1(2.0)=v_{1x}t=(25)(2.0)=50 m$

While object 2 is moving in the horizontal plane with velocity

$v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s$

So its horizontal position at t = 2.0 s is

$x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m$

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that numerically e

Neko [114]

Answer:

the time at which it passes through the equilibrum position is:

t = 0.1 second

Explanation:

given

w= 4pounds

k(spring constant) = 2lb/ft

g(gravitational constant) = 10m/s² = 32ft/s²

β(initial point above equilibrum) = 1

velocity = 14ft/s

attached is an image showing the calculations, because some of the parameters aren't convenient to type.

3 0

3 years ago

Describe what a reactant is

ohaa [14]

Answer:

the element that takes part in a chemical reactions

there are usually two or more reactants in chemical reactions

except decomposition reactions

4 0

4 years ago

Read 2 more answers