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3 years ago

Enter the following expression in the answer box below: 2gλ3m−−−−√, where λ is the lowercase Greek letter lambda.

Physics

1 answer:

atroni [7]3 years ago

7 0

Answer:

$\sqrt{\dfrac{2g\lambda^3}{m}}$

Explanation:

We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.

$\boxed{\sqrt{\dfrac{2g\lambda^3}{m}}}$

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From her bedroom window a girl drops a water-filled balloon to the ground, 4.60 m below. If the balloon is released from rest, h

anygoal [31]

Known variables
d=4.6m
initial velocity=0m/s
downward acceleration=-9.8m/s2

d=1/2gt2
4.6=1/2 -9.8 t2
t=0.93s

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4 years ago

A boat, whose speed in still water is 8.0 m/s, crosses a river with a current of 6.0 m/s. If the boat heads perpendicular to the

kirill [66]

Answer: D

Rs = 10.0 m/s

The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s

Explanation:

Since the boat is moving perpendicular to the current of the river, the speed of the boat has two components.

i. 8.0m/s in the direction perpendicular to the current

ii. 6.0m/s in the direction of the current.

So, the resultant speed can be derived by using the equation;

Rs = √(Rx^2 + Ry^2)

Taking

Ry = 8.0m/s

Rx = 6.0m/s

Substituting into the equation, we have;

Rs = √(6.0^2 + 8.0^2)

Rs = √(36+64) = √100

Rs = 10.0 m/s

The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s

3 0

4 years ago

Read 2 more answers

A box of mass m1 rests on a smooth, horizontal floor next to a box of mass m2. Suppose the force of 20.0 N pushes on two boxes o

meriva

Answer:

Explanation:

Given

acceleration of system a =1.2 m/s^2

Normal Force N=4.45 N

Force exerted F=20 N

Thus

$F=(m_1+m_2)a$

$\frac{20}{1.2}=m_1+m_2$

$16.67=m_1+m_2$ -------1

Normal reaction $N=m_2a$

$4.45=m_2\times 1.2$

$m_2=3.70 kg$

therefore $m_1=16.67-3.70$

$m_1=12.96 kg$

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3 years ago

Read 2 more answers

A bob attached to a string of length L = 1.25 m, initially found at the equilibrium

malfutka [58]

The maximum displacement angle of the bob is 13⁰.

The given parameters;

<em>Length of the pendulum, L = 1.25 m</em>
<em>Initial velocity of the bob, v = 0.8 m/s</em>

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m$

The maximum displacement angle is calculated as follows;

$cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0$

Thus, the maximum displacement angle of the bob is 13⁰.

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3 years ago

Choose the false statement concerning parallel circuits.

alisha [4.7K]

Current can vary in different branches of a circuit

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3 years ago