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PilotLPTM [1.2K]
3 years ago
10

A tall tree has more chances to break during the strom why?​

Physics
2 answers:
jolli1 [7]3 years ago
7 0

Answer:

tall tree are more susceptible to windthrow

S_A_V [24]3 years ago
6 0

Explanation:

Trees, because of their height, are natural lightning rods. Electricity seeks the path of least resistance, and the moisture inside a tree is a much better conductor than air. This means a tree provides the preferred path for lightning to reach the ground.

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The motion of a car is represented as shown. The average
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What part of an electric circuit are the wires that carry the electric current
bezimeni [28]

Answer:

okay so, An electric circuit is a closed loop through which current can  actually flow. All electric circuits must have a voltage source, for example a battery, and a conductor, which is usually  a wire.

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8 0
4 years ago
Using the equation T=2π√L/g solve for L
weqwewe [10]

You said  T = 2 · π · √(L / g)

(I think that's the formula for the full-swing period of a simple pendulum.)

Divide each side by  2π :  T/2π = √(L / g)

Square each side:  (T/2π)²  =  L / g

Multiply each side by ' g ' :  L = g · (T/2π)²

4 0
3 years ago
A car is coasting backwards downhill at a speed of 3.0 m/s when the driver gets the engine started. After 2.5
astra-53 [7]

Answer:

0.6m/s²

Explanation:

Given parameters:

Initial velocity  = 3m/s

Final velocity  = 4.5m/s

Time taken  = 2.4s

Unknown:

Average acceleration  = ?

Solution:

To solve this problem, we use the expression:

 Acceleration  = \frac{v - u}{t}  

v is the final velocity

u is the initial velocity

t is the time taken

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4 0
3 years ago
As in Problem A above, a block of mass M starts from rest and is pushed up a frictionless ramp inclined at an angle θ above the
Tasya [4]

Answer:

A(i)

The solution to this question is shown on the second uploaded image

A(ii)

The final speed is v = \sqrt{2L (\frac{F}{M} - gsin\theta )}

B

The block speed after a distance L  is  v= \sqrt{2L (\frac{F}{M} -gsin \theta )}

Explanation:

From the question  

The net force i the x-direction is mathematically represented as

               F_{net} = Ma

From the the diagram in the second uploaded image we see that

             F_{net} = F - Mgsin \theta

Therefore

              F- Mgsin\theta = Ma

Making a the subject

              a = \frac{F}{M} - gsin\theta

Applying the law of motion

                v^2 =u^2 + 2as

where  u = 0 m/s and s =L

       v^2 = 0 + 2(\frac{F}{M} - gsin \theta )L

=>   v = \sqrt{2L (\frac{F}{M} - gsin\theta )}

According to Energy conservation law and work theorem

           Workdone by F + Workdone by gravity = change in kinetic energy

Mathematically this is given as

                F * L - (mgsin \theta)L = \frac{1}{2} M (v^2-u^2)

   Since u = 0 m/s

                    L (\frac{F}{M} - gsin \theta ) = \frac{1}{2} v^2

                  v= \sqrt{2L (\frac{F}{M} -gsin \theta )}

5 0
3 years ago
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