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Maurinko [17]
2 years ago
6

0.

Physics
1 answer:
larisa [96]2 years ago
4 0

Answer:21

Explanation:

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Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m
trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ =  30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T:  cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction  x-axis :

∑Fx = m₁*ax   ,ax  =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470  -258.77 -300*a =T

T= 1211.23-300*a   Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay   ,ay  =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T:  cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends  vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) =  Equation (2) = T

1211.23-300*a= 980  + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends  vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0
3 years ago
Scientific investigations
snow_tiger [21]
I don’t know what you are trying to ask complete my but here are the steps to a scientific investigation i hope this helps you

6 0
3 years ago
HELP WILL MEDAL!!!! HELP
steposvetlana [31]
Here try this. The pic is the answer

8 0
3 years ago
In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev
Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

6 0
3 years ago
Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the
Dafna11 [192]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Heat capacity of body 1 :

\qquad \sf  \dashrightarrow \:m_1s_1

Heat capacity of body 2 :

\qquad \sf  \dashrightarrow \:m_2s_2

it's given that, the the head capacities of both the objects are equal. I.e

\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2

\qquad \sf  \dashrightarrow \:m_1 =  \dfrac{m_2s_2}{s_1}

Now, consider specific heat of composite body be s'

According to given relation :

\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }

[ since, m_2s_2 = m_1s_1 ]

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{  \cancel{m_2}(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 s_2}{  (\frac{s_2 + s_1}{s_1} )}

\qquad \sf  \dashrightarrow \: s' =  \dfrac{2s_1s_2}{s_1 + s_2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

6 0
3 years ago
Read 2 more answers
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