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Novosadov [1.4K]
3 years ago
15

What is the temperature inside of a tree?

Physics
1 answer:
Oliga [24]3 years ago
6 0
It’s 21 c it all on the weather outside but most of the time it’s on 21
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You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y
Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is  55 mi/h

so if he drives all the to the meeting with max speed then it takes =\frac{450}{55}=8.181 h

and total allowable time is 10.8

Therefore longest time he can spend over dinner is 10.8-8.181 \approx 2.6 hours

5 0
3 years ago
Which shows the order of mechanical advantage from least to greatest?
Eduardwww [97]
What do you mean? I'm confused... You need to put the rest  of the question

3 0
3 years ago
A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?
Tcecarenko [31]

magnitude of the net force = mass x acceleraton

                                             = 22 x 2.3

                                             =50.6 N

7 0
3 years ago
The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.
klio [65]
The relationship between frequency and wavelength for an electromagnetic wave is
c=f \lambda
where
f is the frequency
\lambda is the wavelength
c=3 \cdot 10^8 m/s is the speed of light.

For the light in our problem, the frequency is f=1.20 \cdot 10^{13} s^{-1}, so its wavelength is (re-arranging the previous formula)
\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}=  2.5 \cdot 10^{-5}m
8 0
3 years ago
At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately
Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

I = 2 kJ/s*m^2

area  on which photons are received is given as

A = 4.80 cm^2 = 4.80 * 10^-4 m^2

now we can find the power received due to sunlight

P = I*A

P = 2* 10^3 * 4.80 * 10^-4

P = 0.96 Watt

now we can say this power is due to photons that strikes on surface of earth

so here we can say

P = N\frac{hc}{\lambda}

given here that

\lambda = 510 nm

0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}

0.96 = N * 3.88 * 10^{-19}

N = \frac{0.96}{3.88*10^{-19}}

N = 2.47 * 10^{18}

so it will strike 2.47 * 10^18 photons on given area per second

3 0
3 years ago
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