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gayaneshka [121]
3 years ago
12

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is

located at 23.5°N and in December when the sub-solar point is at ________ latitude.
Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is located at 23.5°N and in December when the sub-solar point is at 23.5°S latitude.

Explanation:

At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.

Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.  

During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees

During winter soliscitse on December 21, the sub-solar point is fixed at)  Tropic of Capricorn.

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A construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity if he travels a distance o
Anastasy [175]

Answer:

W = 0

Explanation:

We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.

The work done by an object is given by :

W=Fd

F = ma

So,

W=mad

m is mass

a is acceleration

d is displacement

The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.

8 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid
bixtya [17]

Answer:

u_{s}=0.56

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

F_{s.max}=\frac{mv^2}{r} \\

Where the r is the radius of merry-go-round and v is the tangential speed

but

F_{s.max}=u_{s}F_{N}=u_{s}mg

So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

Substitute the given values

So

u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56

6 0
3 years ago
A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the
adelina 88 [10]

Answer:

maximum possible velocity = \sqrt{ugR}

Explanation:

centripetal acceleration when the  car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration \frac{mv^{2}}{r}

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

thereforeumg\geq \frac{mv^{2}}{R}

therefore maximum possible velocity = \sqrt{ugR}

6 0
3 years ago
Hello, I wanted an answer from a mathematician. The number 1.04 is closer to the number 1, 2, 1.25 or 1.5.
babymother [125]

Answer:

Explanation:

Of the 4 numbers given, the answer is 1 or A

If you take the absolute value of abs(1 - 1.04) you get 0.04.

(2 - 1.04) = 0.96

1.25 - 1.04 = .21

1.5 - 1.04 = 0.46

The last three are all larger than 0.04

Note: absolute value means the positive difference between 2 numbers (even  though it is negative). If it is negative, absolute value makes it positive.

3 0
2 years ago
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
2 years ago
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