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erica [24]
3 years ago
5

PLEASE HELP!I need help with this

Physics
1 answer:
melisa1 [442]3 years ago
3 0

A)

At time = .003 hr

B)

90* (3*10^-3) = .27km

C)

.005 hr

D)

Car A = 150 * (.005) = .75km

Car B = (90 * (.005) = .45 km

E)

Car A = 150 * (0.008) = 1.2 km

Car B = 90 * (0.008) = 0.72 km

Car A is ahead



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A car is traveling with 90km/hr and another car with 20m/s in opposite direction. Calculate the relative velocity.
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90 km/h is 25 m/s
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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

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2 years ago
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A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

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b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

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Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
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Answer:

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⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

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Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

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Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

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3 years ago
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