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koban [17]
3 years ago
15

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

Physics
1 answer:
babymother [125]3 years ago
4 0

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

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A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

\phi = 1.89 \times 10^{-7}t

(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0
3 years ago
Benita is studying the erosion of soil after a heavy rainfall. She observes that rainwater washes away very little soil from are
yanalaym [24]

Answer:

yes bc it washed it away so

Explanation:

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4 0
2 years ago
Calculate the root mean square velocity of nitrogen molecules at 25 C.
iragen [17]
 root mean square<span>= square root of ( 3RT/M) 

 R = 8.314 J/K/mole 
T = 25 + 273 = 298 K 
M = molecular mas of N2 in kg = 28 X 10^-3 kg 

put values... 

</span><span> root mean square</span> = square root of ( 3 X 8.314 X 298/28 X 10^-3)
= square root of ( 265454.143)
= 515.2 m/s
so option A is right
hope this helps
4 0
3 years ago
See attachment for question
zloy xaker [14]

The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.

The ratio of the calculated and accepted value of the Mercury's mass is 0.95.

<h3>What is mass?</h3>

Mass is the amount of matter present in the object.

The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.

Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.

The acceleration due to gravity is related with mass as

g = GM/R²

Substitute the values, we have

3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³

M = 2.2016 x 10¹³ /  6.67 x 10⁻¹¹

M = 0.3152 x 10²⁴ kg

Thus, the mercury's mass is  0.3152 x 10²⁴ kg.

(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg

Ratio of the value of mass calculated and accepted is

Mcalc/M accep =  0.3152 x 10²⁴ kg / 3.301 x 10²³ kg

                          = 0.95

Thus, the ratio is 0.95

Learn more about mass.

brainly.com/question/19694949

#SPJ1

8 0
1 year ago
Is the strange solid/liquid changing behavior of Oobleck the SAME as an ice cube melting?
Dmitrij [34]
I don’t think so because an ice cube melting needs heat and relies on temp while ooblecks transition from solid to quickly depends on force and speed
6 0
2 years ago
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