Of the following statement: Choose the Variable<br><br> X = 4 m

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

3 years ago

Vascular resistance is the friction force that opposes blood flow in a blood vessel. what two factors are the most important in

makvit [3.9K]

The two most important factors are Systematic Vascular Resistance as and Vascular Resistance.

6 0

3 years ago

Which has the most kinetic energy from its motion?

victus00 [196]

Answer:

option c

Explanation:

K.E=1/2mv(squared)

K.E=1/2x900x80x80

K.E=450x6400=2880000J

convert to KJ=2880000/1000=2880KJ

6 0

3 years ago

Plz do it all i will give brainlest and thanks to best answer do it right plz

kirill115 [55]

The answer is D

Explanation:

soundwaves can travel through water aka a liquid but secondary waves cant

7 0

2 years ago

An object having a mass of 11.0 g and a charge of 8.00 ✕ 10-5 C is placed in an electric field E with Ex = 5.70 ✕ 103 N/C, Ey =

dezoksy [38]

Answer : $F_{x} = 45.6\times10^{-2}\ N$ , $F_{y} = 3040\times10^{-5} N$ and $F_{z} = 0\ N$

Explanation :

Given that,

Charge of the object q = $8.00\times 10^{-5}\ C$

Electric field in x-direction $E_{x} = 5.70\times10^{3}\ N/C$

Electric field in y- direction $E_{y} = 380\ N/C$

Electric field in z - direction $E_{z} = 0$

Now, using formula

$F = q E$

Now, the force on the object in x- direction

$F_{x} = 8.00\times10^{-5} C \times5.70\times10^{3} N/C$

$F_{x} = 45.6\times10^{-2}\ N$

The force on the object in y- direction

$F_{y} = 8.00\times10^{-5}\ C\times 380\ N/C$

$F_{y} = 3040\times10^{-5} N$

The force on the object in z- direction

$F_{z} = 8.00\times10^{-5}\ C\times 0$

$F_{z} = 0\ N$

Hence, this is the required solution.

8 0

3 years ago

Of the following statement: Choose the Variable X = 4 m