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3 years ago

14

Pls help me with this question

1 answer:

Arisa [49]3 years ago

4 0

Answer:

dam 15 marks for that question that's ez marks there

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Ground infrared telescopes Versailles at high altitude and dry climate. What explains the location for those infrared telescopes

Natali5045456 [20]

Answer:

High altitude

Explanation:

Infrared rays are absorbed by water vapor in the lower parts of our atmosphere such as near the sea level.Infrared telescopes positioned on high mountains can observe the cosmos at a wavelength near-infrared. In this case, this telescope is positioned on a mountain top where the altitude is high.

3 0

3 years ago

An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f

Galina-37 [17]

Answer:

T = 17649.03 N = 17.65 KN

Explanation:

The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

$T = W_A = m(g+a)\\$

where,

T = Tension in cable = ?

$W_A$ = Apparent weight

m = mass = 1603 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

$T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\$

<u>T = 17649.03 N = 17.65 KN</u>

8 0

3 years ago

Suppose that 100g of aluminum at 23 degrees C is added to 50g of water at 95 degrees C. Determine the equilibrium temperature of

blagie [28]

Heat capacities of both the substances be given in such situations

5 0

3 years ago

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When calculating acceleration, to find the change in velocity, you subtract the ____________________ velocity from the _________

Harman [31]

Answer:

Smaller velocity from the bigger one

Explanation:

When one calculates velocity they subtract the smaller velocity from the bigger one.

HOPE IT HELPED

3 0

3 years ago

Read 2 more answers

A cannon shoots an artillery shell with a initial velocity of 400 meters/second at an indirect fire angle of 60 on level ground

notsponge [240]

Answer: 14139.19 m

Explanation:

This situation is related to parabolic motion and can be solved using the following equations:

$x=V_{o}cos \theta t$ (1)

$y=y_{o}+V_{o} sin \theta t+\frac{g}{2}t^{2}$ (2)

Where:

$x$ is the horizontal distance (where the artillery shell lands)

$V_{o}=400 m/s$ is the initial velocity

$\theta=60\°$ is the angle

$t$ is the time

$y=0 m$ is the final height

$y_{o}=0 m$ is the initial height

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

So, let's begin by isolating $t$ from (2):

$0=V_{o} sin \theta t+\frac{g}{2}t^{2}$ (3)

$t=-\frac{2 V_{o}sin \theta}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (-\frac{2 V_{o}sin \theta}{g})$ (5)

Rewriting (5) and taking into account $sin(2\theta)=2 sin \theta cos \theta$ :

$x=-\frac{V_{o}^{2}sin(2\theta)}{g}$ (6)

$x=-\frac{(400 m/s)^{2}sin(2(60\°))}{-9.8 m/s^{2}}$ (7)

Finally:

$x=14139.19 m$

6 0

3 years ago