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Tju [1.3M]
3 years ago
7

if you have a stick magnet and a u magnet plus a frigde magnet what way will it pull and which one is the strongest if it does n

ot pull any way?
Physics
1 answer:
SpyIntel [72]3 years ago
7 0

Answer:

the frige magnet because it is bigger and larger

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Gradually slow down to a safe speed well before entering the curve.
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Sergeu [11.5K]

The vertical component of the initial velocity is v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is v_0_x = \frac{x}{t}

The horizontal displacement when the object reaches maximum height is X = \frac{xy}{gt^2} + \frac{x}{2}

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

Learn more here: brainly.com/question/20689870

8 0
3 years ago
A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?
rosijanka [135]

Answer:

a

     l_o  =52 \  m

b

      l = 37.13 \ LY

Explanation:

From the question we are told that

    The  speed of the spaceship is  v  =  0.800c

    Here  c is the speed of light with value  c =  3.0*10^{8} \ m/s

    The  length is  l = 31.2 \  m

     The  distance of the star for earth is d = 145 \  light \  years

     The  speed is v_s = 2.90 *10^{8}

     

Generally the from the length contraction equation we have that

       l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}

Now the when at rest the length is  l_o

So  

      l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }

      l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }

      l_o=52 \  m

Considering b  

  Applying above equation

            l  =l_o \sqrt{1 -  [\frac{v}{c } ]}

Here l_o  =145 \  LY(light \ years )

So

           l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }

            l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }

            l = 37.13 \ LY

4 0
3 years ago
PLEASE HELP ASAP!!!
kirza4 [7]

Answer:

0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4

6 0
3 years ago
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