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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

For installation with a 25-kVA, 3-phase transformer, a 440-volt primary, and a 120-volt secondary. Calculate the maximum overcur

andre [41]

Answer:

41.053 A

Explanation:

given,

three phase kVA = 25-kVA

voltage = 440 Volt

current = ?

To determine three phase kVA when volts and amperes are know

three phase kVA = 1.73 x V x I

$I = \dfrac{three\ phase\ kVA}{1.73 \times V}$

$I = \dfrac{25000}{1.73 \times 440}$

I = 32.84 A

the maximum overcurrent protection value is equal to

= 125 % of I

= 1.25 x 32.84

= 41.053 A

4 0

3 years ago

What resistance would produce a current of 200A with a potiential difference of 200V?

viktelen [127]

Aw, I hate physics, is this on Apex?

Resistance can be calculated with the information given in the question.
Equation for Resistance: R = V/I
V (voltage) = 200 Volts
I (current) = 200 Amps

So 200 divided by 200 = freaking 1

Answer: R = 1 (ohms)

Hope this Helps!

5 0

3 years ago

Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the app

Musya8 [376]

Answer:

7.468 kN

Explanation:

Here the force is given in Newton

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

Tera = 10¹²

The number is 7468.0

Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.

1 kilonewton = 1000 Newton

$1\ Newton=\frac{1}{1000}\ kilonewton$

$\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton$

So, 7468 N = 7.468 kN

7 0

3 years ago

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

6 0

2 years ago