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2 years ago

The amount of intensity and duration of sunlight striking earth vary with latitude

Physics

2 answers:

Natali5045456 [20]2 years ago

8 0

Its true for sure

hope this helps

Marina CMI [18]2 years ago

4 0

true is the correct answer hope it helps

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

2 years ago

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you

soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

$|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s$

The interaction time to avoid that the water balloon breaks is 0.029s

5 0

3 years ago

How can spectroscopy and infrared technology be useful in space?

olchik [2.2K]

Answer:

Explanation:

Spectroscopy measures electromagnetic radiation.

3 0

2 years ago

Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its prev

masha68 [24]

Answer:

The speed in the smaller section is $43.2\,\frac{m}{s}$

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

$\Delta Q=0$ (1)

With Q the flux of water that is $Av$ with A the cross section area and v the velocity, so by (1):

$A_{2}v_{2}-A_{1}v_{1}=0$

subscript 2 is for the smaller section and 1 for the larger section, solving for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}$ (2)

The cross section areas of the pipe are:

$A_{1}=\frac{\pi}{4}d_{1}^{2}$

$A_{2}=\frac{\pi}{4}d_{2}^{2}$

but the problem states that the diameter decreases 86% so $d_{2}=0.86d_{1}$ , using this on (2):

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1}$

$v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s}$

5 0

2 years ago

iogann1982 [59]

Answer:

6 m/s

Explanation:

used omni wavelength calculater

3 0

2 years ago