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Oduvanchick [21]
3 years ago
11

If Earth's clouds did not trap and release heat, how would most climate zones on Earth be affected?

Physics
1 answer:
Morgarella [4.7K]3 years ago
6 0
Tempts will go higher
You might be interested in
A 100 kilowatt bulb burns for 3 hours how much energy did it use
belka [17]

Answer:

1.08×10⁹ J

Explanation:

From the question given above, the following data were obtained:

Power (P) = 100 kilowatt

Time (t) = 3 hours

Energy (E) =?

Next, we shall convert 100 kilowatt to Watts. This can be obtained as follow:

1 KW = 1000 W

Therefore,

100 KW = 100 KW × 1000 W / 1 KW

100 KW = 100000 W

Thus, 100 KW is equivalent to 100000 W.

Next, we shall convert 3 hrs to second (s). This can be obtained as follow:

1 h = 3600 s

Therefore,

3 h = 3 h × 3600 s / 1 h

3 h = 10800 s

Thus, 3 h is equivalent to 10800 s.

Finally, we shall determine the amount of energy consumed as follow:

Power (P) = 100000 W

Time (t) = 10800 s

Energy (E) =?

P = E/t

100000 = E / 10800

Cross multiply

E = 100000 × 10800

E = 1.08×10⁹ J

Therefore, 1.08×10⁹ J of energy was consumed.

4 0
3 years ago
How does the digestive system work with the blood vessels of the circulatory system?
igor_vitrenko [27]
Inhales, digestive system
7 0
3 years ago
Read 2 more answers
Best answer get BRAINLEIST
Lilit [14]
I think it's the last one.



Hope I helped
7 0
4 years ago
What measures the mount of displacement in a longitudinal esbelta?
Travka [436]

<span>The amplitude. It is the displacement at a peak.</span>
8 0
3 years ago
An electron is moving at a speed of 2.20 ✕ 104 m/s in a circular path of radius of 4.3 cm inside a solenoid. The magnetic field
hodyreva [135]

Answer:

a) 2.90*10^-6 T

b) 0.092A

Explanation:

a) The magnitude of the magnetic field is given by the formula for the calculation of B when it makes an electron moves in a circular motion:

B=\frac{m_ev}{qR}

me: mass of the electron = 9.1*10^{-31}kg

q: charge of the electron = 1.6*10^{-19}C

R: radius of the circular path = 4.3cm=0.043m

v: speed of the electron = 2.20*10^4 m/s

By replacing all these values you obtain:

B=\frac{(9.1*10^{-31}kg)(2.20*10^4 m/s)}{(1.6*10^{-19}C)(0.043m)}=2.90*10^{-6}T=2.9\mu T

b) The current in the solenoid is given by:

I=\frac{B}{\mu_0 N}=\frac{2.90*10^{-6}T}{(4\pi*10^{-7}T/A)(25)}=0.092A=92mA

8 0
3 years ago
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