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nekit [7.7K]
3 years ago
10

How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would

appear to remain in the same position in the sky)? Assume the radius of the earth at the equator is 3960 miles.
Physics
1 answer:
professor190 [17]3 years ago
5 0

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

Circumference=2\pi R

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

v=\frac{Distance}{Time}

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

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Why wheat flour is usually passed near a magnet before being packed?
densk [106]
Becasue steel particles may contaminate the flour.
8 0
3 years ago
The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizon
kirza4 [7]

Answer:

A) 3.79 m/s  B) 1.33 m

Explanation:

A)

  • Horizontal movement:
  • Once in the air, no forces act on the froghopper, so it keeps moving with the same initial horizontal speed.
  • This horizontal component, is the projection of the velocity vector on the horizontal direction (x-axis):

        v_{ox} = v_{o} *cos (58.0 deg)

  • The horizontal displacement can be simply calculated as follows:

        x = v_{ox} *t

  • Vertical movement:
  • As the vertical and horizontal are independent each other (due to they are perpendicular, so there is no projection of one movement on the other), in the vertical direction, all happens as if would be a body thrown upward with a given initial vertical velocity.
  • This velocity can be found as the projection of the velocity vector on the vertical direction (y-axis):

        v_{oy} = v_{o} *sin (58.0 deg) (1)

  • Once in the air, the gravity will cause that the froghopper be slow down, till it reaches to the maximum height, where it will come momentarily to an stop.
  • In that moment, we can apply the following kinematic equation:

        v_{fy} ^{2} -v_{oy} ^{2} = 2*g*h_{max}

  • where vfy = 0, g = -9.8m/s2, hmax = 52.7 cm= 0.527 m
  • Replacing by the givens, we can solve for voy:

        v_{oy} =\sqrt{2*g*h_{max}} = \sqrt{2*9.8m/s2*0.527m} =3.21 m/s

  • From the equation (1), we can solve for the magnitude of the initial velocity, v₀:

        v_{o} = \frac{v_{oy}}{sin 58.0} =\frac{3.21m/s}{0.848} = 3.79 m/s

B)

  • With the value of the magnitude of the initial velocity, we can find the horizontal component vox, as follows:

        v_{ox} = v_{o} *cos (58.0 deg) =\\  \\ 3.79 m/s * cos (58.0deg) = 2.01 m/s

  • In order to know the horizontal distance travelled, we need to find the time that the insect was in the air.
  • We can use the equation for the vertical displacement, replacing this value by 0, as follows:

       y = 0 = v_{oy} *t -\frac{1}{2} * g *t^{2}

  • Replacing by  the givens, and rearranging terms, we can solve for t:

        t_{air} =\frac{2*v_{oy} }{g} = \frac{2*3.21 m/s}{9.8 m/s} = 0.66 s

  • Finally, we find the horizontal displacement, as follows:

       x_{max}  = v_{ox} *t_{air} = 2.01 m/s * 0.66 s \\ \\ x_{max} = 1.33 m

  • The horizontal distance covered by  the froghopper was 1.33 m.
4 0
3 years ago
At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi
Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0
3 years ago
A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of
SVEN [57.7K]

Answer:

Acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

T=2\pi \sqrt{\frac{l}{g}}

2=2\times 3.14 \sqrt{\frac{2}{g}}

0.3184= \sqrt{\frac{2}{g}}

Squaring both side

0.1014= {\frac{2}{g}}

g=19.71=20m/sec^2

So acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer.

8 0
3 years ago
A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m
Anit [1.1K]
 <span>(a) 

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get: 

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component. 

(b) 
To much the same end do we derive the vertical component: 

18.0*sin37.5 = v_y = 10.96 ms^-1 

Which we then divide by acceleration, a_y, to derive the time till maximal displacement, 

10.96/9.8 = 1.12 s 

Finally, doubling this value should yield the particle's total time with r_y > 0 

<span>2.24 s

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span></span>
6 0
3 years ago
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