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Zina [86]
2 years ago
14

How do i figure this out ? Anyone help me please

Physics
1 answer:
kondor19780726 [428]2 years ago
6 0

Answer:

are you on instagram follow me lewa_boy

Explanation:

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A spider of mass mm is swinging back and forth at the end of a strand of silk of length LL. During the spider's swing the strand
krok68 [10]

Answer:

The speed of the spider is v = (2g*L*(1-cosθ))^1/2

Explanation:

using the energy conservation equation we have to:

Ek1 + Ep1 = Ek2 + Ep2

where

Ek1 = kinetic energy = 0

Ep1 = potential energy = m*g*L*cosθ

Ek2 = (m*v^2)/2

Ep2 = m*g*L

Replacing, we have:

0 - m*g*L*cosθ = (m*v^2)/2 - m*g*L

(m*v^2)/2 = m*g*L*(1-cosθ)

v^2 = 2g*L*(1-cosθ)

v = (2g*L*(1-cosθ))^1/2

4 0
3 years ago
If you're driving 55 MPH and you suddenly need to stop, how many feet will you travel before the car comes to a stop
lys-0071 [83]
ANSWER: 170 Feet
REASON: with good breaks and a dry road your car should stop and skip 170 feet, with perception and reaction time of stopping you should stop within 170 feet
5 0
2 years ago
I WILL GIVE BRAINLIEST!!!
astra-53 [7]

Answer:

is that high school work??? cause I don't know it and I'm about to go to high school

7 0
3 years ago
Tomorrow i have a science test, could someone please fill me in on the water cycle?
Lorico [155]

the cycle of processes by which water circulates between the earth's oceans, atmosphere and land involving precipitation as rain and snow drainage in streams and rivers and return to the atmosphere by evaporation and transpiration.

Hope this gives you a little bit more information!

5 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
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