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weeeeeb [17]
2 years ago
5

(Thought question) At noon On June 21 will the shadow length of 42 degrees north latitude be shorter or longer than it is on Mar

ch 21? And, what is your reason for saying it will be either shorter or longer?
Physics
1 answer:
g100num [7]2 years ago
5 0

Answer: Shorter

Explanation: Shadow is formed when an light source is obstructed by an opaque object. The closer the source, shorter is the length of the shadow. In fact, when the source is exactly overhead, no shadow of the object is formed.

June 21 marks the Summer solstice which means the Sun passes directly overhead Tropic of cancer (23.5° N) at noon. March 21 marks the equinox which means sun passes directly overhead equator (0°).

Shadow length of an object at 42° Northern latitude will be shorter on June 21 because the Sun will be closer to this latitude as compared to March 21.

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prove that the rate of heat production in each of the two resistors connected in parallel are inversely proportional to the resi
Katyanochek1 [597]
I believe that the answer to the question provided above is that with increase in resistance provided with constant current, Power dissipated will be lessen since power loss is high. Low power dissipation has low heat production.
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
5 0
3 years ago
A projectile of mass m is launched with an initial velocity vector v i making an angle θ with the horizontal as shown below. The
sergeinik [125]
Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be: 
<span>L = R * m * vi * cos(90 - theta) </span>

<span>cos(90 - theta) is just sin(theta) </span>
<span>and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g </span>

<span>so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g </span>

<span>We can combine the two vi terms and get: </span>

<span>L = vi^3 * m * sin(theta) * sin(2*theta) / g </span>

<span>What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. </span>

<span>Now, for the first part... </span>

<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>

<span>Okay, so let's get back to what we know: </span>

<span>L = d * m * v * cos(phi) </span>

<span>where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... </span>

<span>m is still m. </span>
<span>v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) </span>
<span>d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g </span>

<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>

<span>The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: </span>

<span>h = 1/2 * (vi * sin(theta))^2 / g </span>

<span>So, there's the rise. So, our *slope* is rise/run, so </span>

<span>slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] </span>

<span>The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: </span>

<span>slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] </span>

<span>(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). </span>

<span>Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... </span>

<span>slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 </span>

<span>Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... </span>

<span>alpha = arctan( tan(theta) / 4 ) </span>

<span>Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha </span>

<span>so, phi = 180 - arctan( tan(theta) / 4 ) </span>

<span>Now, we go back to our original formula and plug it ALL in... </span>

<span>L = d * m * v * cos(phi) </span>

<span>becomes... </span>

<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] </span>

<span>Now, cos(180 - something) = cos(something), so we can simplify a little bit... </span>

<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ] </span>
3 0
2 years ago
Read 2 more answers
A certain car can accelerate from 0 to 60 mph in 7.9 s. What is the car's average acceleration in mph/s?
Anna007 [38]

Answer:

<em>a = 7.6\ mph/s</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

v_f=v_o+at

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

\displaystyle a=\frac{v_f-v_o}{t}

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

\displaystyle a=\frac{60 \ mph-0}{7.9}

a = 7.6\ mph/s

7 0
3 years ago
What happens when the sun emits more energy than normal
xxTIMURxx [149]

Answer:

When the Sun emits more amount of energy than normal, "Solar flares and sunspots" occur, increasing temperature of Earth. Explanation: The Earth's temperature is governed by many factors. One of these factors is 'Solar flare'.

4 0
3 years ago
Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizon
Tems11 [23]

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

  • Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
  • Fperpendicular = F*cosα =  300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

<u><em>The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.</em></u>

6 0
3 years ago
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