The invisible line that passes through the North Pole and the South<br> Pole is called Earth's what

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

bija089 [108]

Answer

Mass m = 78 kg

Vertical height in each stage h = 11 m

(a).

Initial speed u = 0

Final speed v = 1.1 m / s

$v^2=u^2 + 2 as$

$1.1^2 = 2 a \times 11$

a = 0.055 m/s²

Work done

$W_a= m g h + \dfrac{1}{2}mv^2$

$W_a= 78\times 9.8 \times 11 + \dfrac{1}{2} 78 \times 1.1^2$

$W_a = 8408.4 + 47.19$

$W_a = 8455.59 J$

(b).Work done

$W_b= mgh$

W_b = 78× 9.8× 11

$W_b= 8408.4 J$

c)

Work done

$W= m g h + \dfrac{1}{2}m(v_f-v_i)^2$

Where V = final speed

= 0

v = 1.1 m / s

for deceleration a = -0.055 m/s²

now,

$F_L = 56 (-0.055+9.8) = 545.72 N$

W_c = 545.75 × 11

$W_c = 6003.25 J$

7 0

3 years ago

A 60-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

Density of the chain = $\rho\ =\ 10\ kg/m.$
Length of the chain = L = 60 m
Acceleration due to gravity = g = 9.81 $m/s^2$

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = $\rho dy$

Work done due to the small element,

$dw\ =\ \rho g (60\ -\ y)dy\\$

Total work done to wind the entire chain = w

$w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J$

part (b)

mass of the block connected to the chain = m = 35 kg

Total work done to wind the chain = work done due to the chain + work done due to the mass

$\therefore W\ =\ w\ +\ mgL\\\Rightarrow W\ =\ 176580\ +\ 35\times 9.81\times 60\\\Rightarrow W\ =\ 176580\ +\ 20601\\\Rightarrow W\ =\ 197381\ J$

4 0

2 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

il63 [147K]

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium ε around .

4 0

3 years ago

A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i

antiseptic1488 [7]

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

= 1000 (0 - 20)

= 1000 (-20)

= -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

4 0

2 years ago

How wood helps in home insulation plz answer fast<br>plz answwweeeeeeerrrrrrr

Usimov [2.4K]

Wood is a poor conductor and therefore a good insulator keeping heat inside

3 0

3 years ago

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The invisible line that passes through the North Pole and the South Pole is called Earth's what