A 30 kg dog runs at a speed of 15 What is the dog's kinetic energy?

Key concepts (Example | scientific method

nekit [7.7K]

Vocabulary should be, I think:
I. Hypothesis
II. Evidence, data
III. Experiment

What is your question exactly?

3 0

3 years ago

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Which type of wave is not a light wave

AlekseyPX

A. RED WAVES IS NOT A LIGHT WAVE BECAUSE THERE IS NOT RED WAVE.

ALSO IF YOU DONT BELIEVE SEARCH IT FROM GOOGLE

4 0

3 years ago

What is short circuit

Marizza181 [45]

Answer:

A short circuit is an electrical circuit that allows

Explanation:

3 0

2 years ago

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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

SpyIntel [72]

Answer:

a) 2.43 m/s

b) 4.83 m/s

c) 0.023 m/s²

Explanation:

a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration

Speed = Distance / Time

$\text{Speed}=\frac{510}{210}=2.43\ m/s$

Velocity of car A is 2.43 m/s

t = Time taken = 210 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 510 m

a = Acceleration

c)

$s=ut+\frac{1}{2}at^2\\\Rightarrow 510=0\times 210+\frac{1}{2}\times a\times 210^2\\\Rightarrow a=\frac{510\times 2}{210^2}\\\Rightarrow a=0.023\ m/s^2$

Acceleration of car B is 0.023 m/s²

b)

$v=u+at\\\Rightarrow v=0+0.023\times 210\\\Rightarrow v=4.83\ m/s$

Final velocity of car B is 4.83 m/s

6 0

3 years ago