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Vaselesa [24]
3 years ago
8

Solve this question please

Physics
2 answers:
mestny [16]3 years ago
4 0

ok so pretty much what you need to do is like ok hmm let me think... right ok. so yeah i gtg

Assoli18 [71]3 years ago
3 0

Answer:

(  \nwarrow^{ + }) \sum \:  F_{{y_{A}}_{2kg}} = F_{{y_{N_A}}_{2kg}}  - m_{{y_{A}}_{2kg}}g \cos( \theta_{{y_{A}}_{2kg}}) = 0 \\ F_{{y_{N_A}}_{2kg}} = m_{{y_{A}}_{2kg}}g \cos( \theta_{{y_{A}}_{2kg}})...(1) \\ (  \nearrow^{ + }) \sum \:  F_{{x_{A}}_{2kg}} = F_{{x_{T}}}  - m_{{y_{A}}_{2kg}}g \sin( \theta_{{y_{A}}_{2kg}}) -F_{{Fr}_{A2kg}}= m_{{y_{A}}_{2kg}}a...(2) \\ \\  (  \nearrow^{ + }) \sum \:  F_{{y_{B}}_{5kg}} = F_{{y_{N_B}}_{2kg}}  - m_{{y_{B}}_{5kg}}g \cos( \theta_{{y_{B}}_{5kg}}) = 0 \\ F_{{y_{N_B}}_{5kg}} = m_{{y_{B}}_{5kg}}g \cos( \theta_{{y_{B}}_{5kg}})...(3) \\ (  \searrow^{ + }) \sum \:  F_{{x_{B}}_{5kg}} =   m_{{y_{B}}_{5kg}}g \sin( \theta_{{y_{B}}_{5kg}})-F_{{x_{T}}} -F_{{Fr}_{B5kg}}= m_{{y_{B}}_{5kg}}a...(4)

Explanation:

All you need to do is plug in the values and solve the simultaneous equations to find the acceleration a, and T.

Always draw the free body diagram...

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Answer: i think its true

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Can the aurora australis be seen from space?
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Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t
BabaBlast [244]

Answer:

-\frac{8\pi}{3}rad/s^2

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

\omega_f^2-\omega_i^2=2\alpha\theta

Where for definition,

\omega_i = \frac{\theta}{t}

The number of revolution (\theta)was given by 20 times, then

\omega_i = \frac{20*2pi}{5}

\omega = 8\pi rad/s

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

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The cook at the end stop to spin, then using our first equation,

0-8\pi = 2\alpha (12\pi)

re-arrange to solve\alpha ,

\alpha = \frac{-8\pi}{2*12\pi}

\alpha = -\frac{8\pi}{3}rad/s^2

We can know find the required time,

\omega_f-\omega_i = \alpha t

Re-arrange to find t, and considering that \omega_f=0

t= \frac{\omega_i}{\alpha}

t=\frac{-8\pi}{-8\pi/3}

t=3s

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6 0
3 years ago
Read 2 more answers
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kozerog [31]

Answer:

h=17m

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s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

to find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is straight line of equation y=9.8 as acceleration is constant:

Velocity graph is given by y=9.8x ( y as velocity and x as time):

Displacement graph is given by y=4.9x^2 ( x as time, y as displacement):

These graphs are only applicable from x=0 to x=1.9 ... ignore the other graph sections

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3 years ago
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aleksandrvk [35]

Answer:

Explanation:

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1. Image formed cannot be projected or focused on a screen.

2. The distance of the object to the mirror is the same as the distance from the image to the mirror.

3. The size of the image formed is the same as the size of the object.

4. The image formed is laterally inverted. That is the right becomes left and vice versa.

5. The image is upright.

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1. Image formed can be projected on a screen.

2, The distance from the image to the mirror is not the same as the distance from the object to the mirror.

3. The size of the image is not the same as the size of the object.

4. Image formed is upside down.

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