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2 years ago

Differentiate between adhesion and cohesion.

Physics

1 answer:

sergiy2304 [10]2 years ago

8 0

<h2>Answer:</h2>

<h2>Cohesion </h2>

Is the attraction that molecules have for others of their same type.

<h2>Adhesion </h2>

Is the attraction that molecules have for others of different type by intermolecular forces.

A good example of both is water that can stick to itself through hydrogen bonds (cohesion) and can also stick to a glass due to adhesion.

So, while Cohesion is the force of attraction between adjacent particles within the same body, Adhesion is the interaction between the surfaces of different bodies.

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Sirens and flashing lights do NOT indicate that

sergij07 [2.7K]

Golf cart is the answer

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2 years ago

Read 2 more answers

A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.6

andriy [413]

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using $A\int_0^{12}\left(B+Cx)dx$

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

$2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}$

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

$m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6$ ,

hence, the mass of the rod is 126.6 g.

7 0

2 years ago

An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only

Pavel [41]

Answer:

vb = 22.13 m/s

Explanation:

ma = 124 kg

mb = 13 kg

vi = 2.10 m/s

According to the property of conservation of momentum, and considering that, initially, both the astronaut and the bag moved together at 2.10 m/s:

$(m_a+m_b)v_i=m_av_a+m_bv_b$

The minimum final velocity of the bag, vb, the will keep the astronaut from drifting away forever occurs when va = 0:

$(124+13)2.10=124*0+13v_b\\v_b=\frac{287.7}{13}\\v_b= 22.13\ m/s$

The minimum final velocity of the bag is 22.13 m/s.

4 0

2 years ago

You travel at a speed of 24 m/s. How much time will it take you to travel 127m (round to two decimal places if needed)

shtirl [24]

Answer: 5.29 seconds

Explanation:

if the speed is 24 meters per second than it would take you 5.29 seconds or 127meters/24m/s to get 5.29s

4 0

2 years ago

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li

erica [24]

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

Stan's speed = 42.9 m/s

Explanation:

<u>Given:</u>

$u_k$ = initial speed of Kathy = 0 m/s

$u_s$ = initial speed of Stan = 0 m/s
$a_k$ = acceleration of Kathy = $4.73\ m/s^2$

$a_s$ = acceleration of Stan = $3.9\ m/s^2$

<u>Assumptions:</u>

$v_k$ = final speed of Kathy when see catches Stan
$v_s$ = final speed of Stan when Kathy catches him
$s_k$ = distance traveled by Kathy to catch Stan
$s_s$ = distance traveled by Stan when Kathy catches him
$t_k$ = time taken by Kathy to catch Stan = $t$
$t_s$ = time interval in which Kathy catches Stan = $t+1$

Part (a):

Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

$\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\$

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

$s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5$

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3$

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9$

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

7 0

2 years ago