A train leaves the station at the 0.0m marker traveling with a constant velocity of 45.0 m/s. How manyseconds later will the train pass the 1500.Om marker? What is the velocity of the train in km/h?
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The train will pass the 1500.0m marker 33.33 seconds after leaving the station. Its velocity is 162.0 km/h.
How to evaluate time the train will pass the 1500.0m marker and it's velocity
To find the time it takes for the train to pass the 1500.0m marker, we can use the formula: time = distance / velocity. Plugging in the values,
time = 1500.0m / 45.0 m/s
time = 33.33 seconds.
To convert the velocity from m/s to km/h, we use the conversion factor: 1 m/s = 3.6 km/h. Therefore,
velocity = 45.0 m/s * 3.6 km/h/m/s
velocity = 162.0 km/h.
In summary, the train will pass the 1500.0m marker approximately 33.33 seconds after departing the station, and its velocity will be 162.0 km/h
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Answer:
119.88 km/h
Explanation:
1500/45=33.3
use a m/s to km/h calculator
put in 33.3 for m/s and you will get 119.88 km/h.
119.88 km/h.
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A snail can move approximately 0.30 meters per minute then it would cover 4.5 meters in 15 minutes.
What is speed?
The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
As given in the problem a snail can move approximately 0.30 meters per minute and we have to calculate the distance covered in 15 minutes.
speed of the snail = 0.30 meters per minute
time = 15 minutes
distance covered by the snail = speed ×time
distance covered by the snail = 0.30 ×15
distance covered by the snail = 4.5 meters
Thus, the distance covered by the snail in the 15 minutes would be 4.5 meters.
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so, distance = 0.30 meters.min x 15 mins
distance = 4.5 meters
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I hope it is helpful for you ...
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The potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V
What will be the potential difference?
It is given that:-
Charge Q = +4.00 mC or
insulating sphereradius R = 5.00 cm = 0.05 m
The formula for the potentialdifference at the surface of the sphere is given by
The formula for the potentialdifference at the center of the sphere
The difference between the potential difference will be
Thus the potential difference between the center of the sphere and the surface of the sphere will be PD=3.6 x 10^8 V
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Answer:
3.6 x 10^8 V
Explanation:
Q = 4 m C = 4 x 10^-3 C
r = 5 cm = 0.05 m
The formula for the potential at the surface is
Vs = K Q / r = (9 x 10^9 x 4 x 10^-3) / 0.05 = 7.2 x 10^8 V
The formula for the potential at the centre is
Vc = 3/2 Vs
Vc = 1.5 x 7.2 x 10^8 V = 10.8 x 10^8 V
The difference in potential is
V = Vc - Vs = 10.8 x 10^8 - 7.2 x 10^8 = 3.6 x 10^8 V
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212*f....... as that kid tried to explain the combustion process, and failed... the process isn't exactly the same for all heat engines... therefore, there has to be a universal heat that of in which has a standard heat of 212 degrees fahrenheit