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3 years ago

7

The length (L) of a simple pendulum is directly proportional to the square of the time period (T). Which type of graph can best

describe this relationship?

2 answers:

iris [78.8K]3 years ago

8 0

L=T^2 so that would be a parabola

forsale [732]3 years ago

6 0

Answer:

L=T^2 so its parabola

Explanation:

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he Hobbits are building a watchtower so they can prepare to battle in case trolls decide to attack them. One Hobbit will always

joja [24]

Answer:

5m/8

Explanation:

Function T gives the time the Hobbits have to prepare for the attack, T(k), in minutes, as a function of troll's distance, k, in meters.

Function V gives visibility from the watchtower, V(m), in meters, as a function of the height of the watchtower, m, in meters.

Therefore, T(V(m)) will give the time the Hobbits have to prepare for the troll attack as a function of the height, m, of the watchtower.

We can input m into function V to obtain the visibility from watchtower, V(m), in meters. Since visibility indicates the distance you can see, this also gives the distance of the trolls. This can then be input into function T to obtain the time that the Hobbits have to prepare for a troll attack.

Let's find T(V(m)) by substituting the formula for V(m) into function T as shown below.

T(V(M))=T(50m)

=50m/80

We can simplify this as follows:

=50m/80

=5m/8

4 0

3 years ago

Pls do question 1 part d). tysm

Aneli [31]

Answer:

c

Explanation:

cuz its informing the length of 5 and weight on 20N

5 0

3 years ago

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

Vesnalui [34]

Ya know what the day you are talking to you about it and then send me the back link you got it and I don’t want

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3 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

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3 years ago

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3 years ago