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Doss [256]
3 years ago
12

A helicopter’s speed increases from 30 m/s to 40 m/s in 5 seconds.

Physics
2 answers:
ANEK [815]3 years ago
6 0
So, acceleration is the change of velocity in time. If no vectors are used, one can use speed:

a = (final speed - start peed)/ time = 10/5= 1 m/s^2,

Notice that the units are m/s^2, or (m/s)/s. That is speed (m/s) per unit time (s)
MariettaO [177]3 years ago
4 0

its 2 meters per sec

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Magma/lava contains a lot of molten iron so that when it erupts from a volcano and cools, the magnetic field of the earth leaves an imprint in it, just like the cooling iron from above. From this imprint we can tell the strength of the magnetic field and also which direction the north and south poles were at the time.
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Use the drop down menu to complete the sentence. Resistance is measured in units called...
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Ohms is the correct answer
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4 years ago
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An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To accomplish this, energy, in th
Harlamova29_29 [7]

Answer:1.816\times 10^{-19} J

Explanation:

Given

E=\frac{hc}{\lambda }

E=2.18\times 10^{-18}(\frac{1}{n_1^2}-\frac{1}{n_2^2})

where h=Planck constant

c=speed of light

E=2.18\times 10^{-18}(\frac{1}{3^2}-\frac{1}{6^2})

E=2.18\times 10^{-18}\times \frac{1}{12}

E=1.816\times 10^{-19} J

7 0
3 years ago
My Notes A container is divided into two equal compartments by a partition. One compartment is initially filled with helium at a
Ostrovityanka [42]

Answer:

final-temperature = T_{f} = 252.51K

Explanation:

we can solve this problem by using the first law of thermodynamics.

    \Delta U= Q-W

Q= heat added

U= internal energy

W= work done by system

                        E_{final}= E_{initial}

<u> C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}    (1)</u>

C_{v}(N_{2})=1.04\frac{KJ}{Kg K}

C_{v}(He)=5.193\frac{KJ}{Kg K}

now

From equation 1

T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}

T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}

T_{f} = 252.51K

4 0
4 years ago
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

T^2=\frac{4\pi}{GM}r^3

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min

the comet takes around 16.63min

8 0
3 years ago
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