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iogann1982 [59]
3 years ago
9

A man is standing on a platform that is connected to a pulley arrangement, as the drawing shows. By pulling upward on the rope w

ith a force the man can raise the platform and himself. The total mass of the man plus the platform is 113 kg. What pulling force should the man apply to create an upward acceleration of 1.20 m/s2?

Physics
1 answer:
leonid [27]3 years ago
8 0

Answer:

The pulling force that the man should apply to create an upward acceleration of 1.20m/s^{2} is P=621.5N

Explanation:

Hi

As it shows in the drawing at the end, we have that the total mass of the man plus the platform is 113 kg, then the force downward W is W=mg=113Kg*9.8m/s^{2}=1107.4N.

Due the man needs to do a pulling force upward capable of lifting himself and the platform with an acceleration of 1.20m/s^{2}, this force should create an acceleration greater than gravity by 1.20m/s^{2}. then a_{up}=g+1.2m/s^{2}=9.8m/s^{2}+1.2m/s^{2}=11m/s^{2}. Therefore the force should be P_{up}=ma_{up}=113kg*11m/s^{2}=1243N.

Finally, as we have a pulley arrangement connected to the platform, it gives the man a mechanical advantage, so he has to do only half of that upward force

, therefore P=\frac{1243N}{2}=621.5N.

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3 0
3 years ago
two small charged objects repel each other with a force 2.53 when separated by a distance 0.11. if the charge on each object is
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2 years ago
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3 years ago
A solenoid consists of 4200 turns of copper wire. The wire has a diameter of 0.200 mm. The solenoid has a diameter of 1.00 cm. W
Stella [2.4K]

Answer:

a. The length of the solenoid wire is approximately 131.95 m

b. The inductance of the solenoid is approximately 2.078 × 10⁻³ H

c. The length of the solenoid is 0.84  m

d. The current after three time constants have elapsed is approximately 456.1 A

Explanation:

The given parameters are;

The number of turns in the solenoid, N = 4,200 turns

The diameter of the wire, d = 0.200 mm

The diameter of the solenoid, D = 1.00 cm

The voltage of the battery connected to the solenoid, V = 12.0 V

The current increase = 155 mA

The time for the increase = 1.50 millisecond

The internal resistance of the battery is negligible

a. The length of wire needed to form the solenoid, l = π·D·N

∴ l = π × 0.01 × 4,200 ≈ 131.95

The length of the solenoid, l ≈ 131.95 m

b. The inductance, 'L', of the solenoid is given as follows;

L = \dfrac{\mu_0 \cdot N^2 \cdot A}{l}

Where;

μ₀ = 12.6 × 10⁻⁷ H/m

N² = 4,200²

A = The cross sectional area of the solenoid = π·D²/4

l = Length of the solenoid = d × N = 0.0002 m × 4,200 = 0.84  m

∴ L = (12.6 × 10⁻⁷ × 4,200² × 0.01² × π/4)/0.84 ≈ 0.002078 = 2.078 × 10⁻³

The inductance, L ≈ 2.078 × 10⁻³ H

c.) The length of the solenoid = d × N = 0.0002 m × 4,200 = 0.84  m

The length of the solenoid = 0.84  m

d. The current after three time constant

 We have;

∈ = -L × di/dt

di/dt = 155 mA/1.5 ms = 103.\overline 3 A/s

∈ = 103.\overline 3 A/s × 2.078 × 10⁻³ H = 0.21472\overline 6 V

We have;

\tau = \dfrac{t}{\left(ln\dfrac{1}{1-\dfrac{Change}{Final-Start} } \right)}

The change in voltage = 0.21472\overline 6 V

The start voltage = 0 V

The final voltage = 12.0 V

t = 1.5 ms = 0.0015 s

We get;

\tau = \dfrac{0.0015}{\left(ln\dfrac{1}{1-\dfrac{0.21472\overline 6}{12-0} } \right)} \approx 8.3076\times 10^{-2}

τ = L/R

Therefore,

R = L/τ =

The resistance = 2.078 × 10⁻³/(8.3076×10⁻²) = 0.0250

The resistance = 0.0250 Ω

I= \dfrac{V}{R} \cdot \left(1 - e^{-\dfrac{t}{\tau} }\right)

Therefore, after three time constants, we have;

∴ I = (12.0/(0.0250)) × (1 - e⁻³) ≈ 456.1

The current after three time constants have elapsed, I ≈ 456.1 A.

3 0
3 years ago
Can someone please give me this answer to this please
Nuetrik [128]
Uuuu I think 0.20 m/s
3 0
3 years ago
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