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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?

1 answer:

SpyIntel [72]3 years ago

7 0

Answer:

the frige magnet because it is bigger and larger

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Keeping in mind the kinetic energy of a moving vehicle, how can a driver best prepare to enter sharp curves in the roadway

777dan777 [17]

Gradually slow down to a safe speed well before entering the curve.

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3 years ago

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Margarita [4]

D) a car speeding up may i have brainliest hope this help

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3 years ago

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