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3 years ago

I need help Mr or ms tutor

Physics

1 answer:

Ronch [10]3 years ago

7 0

Explanation:

Height is the x-axis, and gravitational potential energy is the y-axis. As the height increases, the gravitational potential energy increases linearly.

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True or false? Magnetic reversals are recorded in the newly formed oceanic crust on BOTH sides of a mid-ocean ridge spreading ce

andrey2020 [161]

answer: false is my answer for thi question

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3 years ago

A racecar driver circles a racetrack with a constant speed of 110 km/hr. Which BEST describes the speed and velocity of the car?

tangare [24]

The answer is B, The speed is constant and the velocity is changing.

8 0

3 years ago

Read 2 more answers

An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0

Reptile [31]

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

$s = ut + \frac{1}{2}at^{2}$

$14 = 0 + \frac{1}{2}a\times 1^{2}$

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

7 0

3 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

3 years ago

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2

sleet_krkn [62]

-- Accelerating at the rate of 8 m/s², Andy's speed
   after 30 seconds is

   (8 m/s²) x (30.0 s) = 240 m/s .

-- His average speed during that time is

   (1/2) (0 + 240 m/s) = 120 m/s .

-- In 30 sec at an average speed of 120 m/s,
   Andy will travel a distance of
   (120 m/s) x (30 sec) = 3,600 m

    = 3.6 km .

"But how ? ! ?", you ask.

How in the world can Andy leave a stop light and then
cover 3.6 km = 2.24 miles in the next 30 seconds ?

The answer is: His acceleration of 8 m/s², or about 0.82 G
is what does it for him.

At that rate of acceleration ...

-- Andy achieves "Zero to 60 mph" in 3.35 seconds,
   and then he keeps accelerating.

-- He hits 100 mph in 5.59 seconds after jumping the light ...
   and then he keeps accelerating.

-- He hits 200 mph in 11.2 seconds after jumping the light ...
   and then he keeps accelerating.

-- After accelerating at 8 m/s² for 30 seconds, Andy and his
   car are moving at 537 miles per hour !
   We really don't know whether he keeps accelerating,
    but we kind of doubt it.

A couple of observations in conclusion:

-- We can't actually calculate his displacement with the information given.
   Displacement is the distance and direction between the starting- and
   ending-points, and we're not told whether Andy maintains a straight line
   during this tense period, or is all over the road, adding great distance
   but not a lot of displacement.

-- It's also likely that sometime during this performance, he is pulled
   over to the side by an alert cop in a traffic-control helicopter, and
   never actually succeeds in accomplishing the given description.

5 0

3 years ago

I need help Mr or ms tutor​

I need help Mr or ms tutor