A bar of soap has a mass of 53.65 grams and its dimensions are 4.6cm by

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

A sphere of volume 1.20×10−3m3 hangs from a cable. When the sphere is completely submerged in water, the tension in the cable is

KATRIN_1 [288]

Answer:

B = 62.9 N

Explanation:

This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid

B = ρ g V

write the equilibrium equation

T + B -W = 0

B = W- T (1)

use the density to write the weight

ρ = m / V

m = ρ V

W = ρ g V

substitute in 1

B = m g -T

B = $\rho_{body}$ g V - T

To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³

calculate

B = 7850 9.8 1.20 10⁻³ - 29.4

B = 92.3 - 29.4

B = 62.9 N

4 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

what has more momentum, a baseball traveling at 4 m/s or a baseball traveling at 16m/s ? and which has more energy ?

Anna007 [38]

We know the formulas for momentum and energy. But they both involve the mass of
the object, and we don't know the mass of the baseball. What can we do ?

It's not a catastrophe. The question only asks which one is bigger. If we're clever,
we can answer that without ever knowing how much the momentum or the energy
actually is. We know that both baseballs have the same mass, so let's just call it
' M ' and not worry about what it really is.

<u>Momentum of anything = (mass) x (speed)</u>
Momentum of the first baseball = (M) x (4 m/s) = 4M
Momentum of the second one = (M) x (16 m/s) = 16M
The second baseball has 4 times as much momentum as the first one has.

<u>Kinetic energy of anything = 1/2 (mass) x (speed squared)</u>
KE of the first baseball = 1/2 (M) x (4 squared) = 8M
KE of the second one = 1/2 (M) x (16 squared) = 128M
The second baseball has 16 times as much kinetic energy as the first one has.

3 0

4 years ago

Which electromagnetic waves have the shortest wavelengths and highest frequencies?

Usimov [2.4K]

Answer: Gamma rays

Explanation: The given waves belong to the electromagnetic spectrum which consists of different electromagnetic radiations arranged in terms of increasing wavelengths or decreasing frequencies.

$E= h\times \nu$