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vovikov84 [41]
3 years ago
10

How do you solve 2A= 4lw+2Lh+6wH for w

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

Answer:

\displaystyle w=\frac{A-Lh}{2l+3H}

Explanation:

Solving Equations by Isolation

We have this equation:

2A= 4lw+2Lh+6wH

Let's solve it for w step by step. First, we flip both sides

4lw+2Lh+6wH =2A

Subtracting 2Lh in both sides

4lw+2Lh+6wH-2Lh =2A-2Lh

Simplifying

4lw+6wH =2A-2Lh

Factoring w

w(4l+6H) =2A-2Lh

Dividing by (4l+6H)

\displaystyle w=\frac{2A-2Lh}{4l+6H}

Simplifying by 2:

\displaystyle \boxed{w=\frac{A-Lh}{2l+3H}}

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Which pair correctly identifies the strongest and weakest fundamental forces,
Kay [80]

Answer: Option B.

Strong nuclear force Gravity.

Explanation:

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7 0
3 years ago
A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r
KIM [24]

Answer:

\omega_f = 1.08 rad/s

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

L_i = L_f

I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f

here we know that

I_o = \frac{1}{2}mR^2

I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2

m_1 = 9.41 kg

r_1 = \frac{4}{5}(1.61) = 1.29 m

m_2 = 21.1 kg

Now from above equation

118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f

118.1(1.73) = 188.45\omega_f

\omega_f = 1.08 rad/s

8 0
3 years ago
You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.
Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s² (downward positive, upward negative)

Time taken by the ball to reach the maximum height

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s

Maximum height

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0
3 years ago
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