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2 years ago

5

A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from its left end. If a force of 50N is appl

ied on it's right end, how much force would need to be applied to the left end?
100N
25N
200N
50N

1 answer:

grigory [225]2 years ago

6 0

The force that would be apply is
100N

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Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

$x_A(t) = \frac{1}{2}a t^2$

where

$a = 6 m/s^2$ is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

$x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2$

where a (acceleration) is the same as traveller A, and

$u_B = 20 m/s$

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get $x_B=0$ , which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

$x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s$

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3 years ago

If you accelerate from rest at 12m/s^2 for 5 seconds, what is your finally velocity?

jeka94

Answer:

$v=60m/s$

Explanation:

Use the Kinematic Equation:

$v=v_o+at$

Plug in what is given and solve

$v=0+(12)(5)$

$v=60m/s$

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2 years ago

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

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$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

(c) energy is being delivered by the battery?

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

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Answer:

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vagabundo [1.1K]

Answer:

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3 years ago

Read 2 more answers