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Nina [5.8K]
3 years ago
11

A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

oes not lie halfway between the two ends. The system is balanced by a support that is 0.30 m from the left end of the plank. How far to the right of the support is the plank
Physics
1 answer:
lesya [120]3 years ago
5 0

Answer:

d_{2}=0.585m

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

T_{1}-T_{2}=0

T_{1}=T_{2}

F_{1}*d_{1}=F_{2}*d_{2}

43N*0.30m=21N*d_{2}

Solve to d2

d_{2}=\frac{41N*0.30m}{21N}

d_{2}=0.585m

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A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
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Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

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= \sqrt { \frac{100*32.2}{200}}

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replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

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3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
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Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

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             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

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Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

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Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

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