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3 years ago

7

How can spectroscopy and infrared technology be useful in space? (5 points)

1 answer:

Mrrafil [7]3 years ago

8 0

Answer:

B

Explanation:

i took the test

You might be interested in

Which of the following describes the effect of a convex mirror on light rays?

natulia [17]

B. It reflects light rays outward.

A convex mirror looks like this ⊂. So when light rays hit the outward curves, it bends the light rays outwards, and backwards, and it would appear to be coming from its center called the focus.

3 0

3 years ago

The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

Brrunno [24]

Answer:

Explanation:

We shall apply concept of impulse to solve the problem .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

4 0

3 years ago

Assertion(A):The distance moved by an object in unit time is called its speed. Reason (R):Faster vehicles have higher speeds. i)

ikadub [295]

Answer:

it's ii) R is correct while A is incorrect

Explanation:

cuz, both statements are correct but the reason is not the correct reason for the assertion,

8 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at

Maurinko [17]

Answer:

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob

Explanation:

Given:

mass of Jacob, $m_j=45\ kg$

velocity of Jacob, $v_j=8.08\ m.s^{-1}$

mass of Ethan, $m_e=31\ kg$

velocity of Ethan, $v_e=10.9\ m.s^{-1}$

Now using the conservation of linear momentum for the case:

(When the two masses in motion combine to form one after the collision then they will move together in the direction of the greater momentum.)

$(m_j+m_e).v=m_j.v_j-m_e.v_e$

$(45+31)\times v=45\times 8.08-31\times 10.9$

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob as it was assumed to be positive.

5 0

3 years ago