(3.561x10^4)(2.89x10^-2)/(2.18x10^11)(5.13161x10^3)=

Why is an absorption spectrum especially useful for astronomers?.

creativ13 [48]

Answer:

It has dark lines in it that allow astronomers to determine what elements are in the star. red-shifted (shifted toward the red end of the light spectrum).

Explanation:

4 0

2 years ago

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 seconds and after 4 seconds, (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hits the ground?

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

Naily [24]

Answer:

700J

Explanation:

8 0

3 years ago

You are asked to design a horizontal curve with a 40-degree central angle (' = 40) for a two-lane road with 11-ft lanes. the des

Alex17521 [72]

The answer is b. False. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h. Give it a try man

3 0

3 years ago

The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.What power is absorbed by the bulb?

Verizon [17]

Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

Power = Voltage × Current

Given:

Voltage = 3 V

Current = 200 mA = 0.200 A

Conversion used : (1 mA = 0.001 A)

Now put all the given values in the above formula, we get:

Power = Voltage × Current

Power = 3V × 0.200 A

Power = 0.600 W

Thus, the power absorbed by the bulb is, 0.600 W

3 0

4 years ago