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3 years ago

CONVERT 5 SOLAR DAYS INTO SECOND

Physics

1 answer:

zhannawk [14.2K]3 years ago

7 0

Answer:

5 solar days is 432,000 seconds

Explanation:

One solar day is the time the meridian of the Sun crosses passes through a specified place.

Due to the elliptical orbit of the Earth round the Sun, one average solar day on Earth is equivalent to 24 hours with some days longer and some days shorter each by a few minutes

Therefore, we have;

5 solar days = 5 × 24 hours × 60 minutes/hour × 60 seconds/minute = 432,000 seconds

5 solar days = 432,000 seconds

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Determination of the energy imparted to matter by radiation is the subject of ______. A. photometry B. magnification C. dosimetr

givi [52]

Answer:

dosimetry

Explanation:

dosimetry is the determination of energy imparted to matter by radiation

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3 years ago

An object of mass 3.07 kg, moving with an initial velocity of 5.07 m/s, collides with and sticks to an object of mass 2.52 kg wi

Alenkasestr [34]

Answer:

This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.

Let's see what conservation of momentum in both directions does ya:

Conservation in the x direction:

Only 1 object here has a momentum in the x direction initally.

m1v1i + 0 = (m1 + m2)(vx)

3.09(5.10) = (3.09 + 2.52)Vx

Vx = 2.81 m/s

Explanation:

Conservation in the y direction:

Again, only 1 object here has initial velocity in the y:

0 + m2v2i = (m1 +m2)Vy

(2.52)(-3.36) = (2.52 + 3.09)Vy

Vy = -1.51 m/s

++++++++++++++++++++

Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)

Vf = √(2.81)^2 + (-1.51)^2

Vf = 3.19 m/s

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4 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago

The velocity of the source is positive if the source is ______________. Note that this equation may not use the sign convention

sveta [45]

Explanation is $^{}$ in a file

bit. $^{}$ ly/3tZxaCQ

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3 years ago

CONVERT 5 SOLAR DAYS INTO SECOND​

CONVERT 5 SOLAR DAYS INTO SECOND