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zimovet [89]
1 year ago
14

If a train travels 500 kilometers from Stockholm to

Physics
1 answer:
o-na [289]1 year ago
4 0

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

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Which statement is false?
podryga [215]

Answer:

D. Current with not flow through a wire if the magnet is outside the coil.

Explanation:

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3 years ago
Why was the geocentric model of the solar system accepted by scientist for many years?
dolphi86 [110]

Answer: Because of the its assumption of earth being at the centre of the universe and also, its parallax argument.

Explanation: Geocentric model is the

theory of the structure of the solar system (or the universe) in which Earth  is assumed to be at the centre of it all. It was generally accepted until the 16th century, after which it was superseded by heliocentric models 

Among many, geocentric model was considered because of its stellar parallax argument.

What is parallax? Parallax is the apparent change in position of objects due to a change in observation location

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6 0
3 years ago
What force is needed to give a 4.5-kg bowling ball an acceleration of 9 m/s2?
Mrac [35]
The correct answer is 40.5 Newtons just finished the quiz and 36.5 was incorrect.
7 0
3 years ago
Read 2 more answers
A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
4 years ago
Keeping the applied voltage constant at approximately 0.70 volts, measure the electric field strength at different points betwee
Vitek1552 [10]

Answer: I observed that the electric field strength is the same at all points between the plates. The value of the field is 70 volts per meter. This is exactly 100 times the applied voltage. The electric field lines point from the positive plate to the negative plate, as the downward arrow on the detector shows.

Explanation:

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3 0
2 years ago
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