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hram777 [196]
4 years ago
13

Please help problems 2a.,2b.,3a.,and 3b.​

Physics
1 answer:
Darina [25.2K]4 years ago
8 0

Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

2a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]

3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]

3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}

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A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 4.79 g coins stacked over the 28.8 cm mark, th
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To solve the exercise, the key concept to be addressed is the Mass Center.

The center of mass of an object is measured as,

X_{cm} = \frac{\sum m_ix_i}{\sum m_i}

X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}

Our values are given by,

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X_{cm} = 38.4cm

Replacing the values in our previous equation we have,

X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}

38.4 = \frac{m_1(50)+2(28.8*4.79)}{m_1+2*4.79}

38.4(m_1+2*4.79)= m_1(50)+2(28.8*4.79)

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