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aleksklad [387]
4 years ago
15

What is the fastest time trial for the first quarter checkpoint? _____ seconds

Physics
2 answers:
Dennis_Churaev [7]4 years ago
6 0

Answer:

What is the fastest time trial for the first quarter checkpoint? 2.02 seconds

What is the slowest time trial for the first quarter checkpoint? 2.15 seconds

What is the range of times measured for this checkpoint? 0.13 seconds

Ivanshal [37]4 years ago
3 0
We don’t know the data, we would have to be given the data to answer these questions
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How do you post a picture
BARSIC [14]
Click the ask a question button or answer button then at the bottom above the keyboard there are two little icons. A camera lens and a Mountain View. If you click the camera lens you can take a picture to post and if you click the Mountain View it will give you the option to post a picture from your photos.
4 0
3 years ago
Find the current flowing through the following circuit
kap26 [50]

Answer:

0.5

Explanation:

calculating the total resistance for the both the parallel and the series resistors then later using the ohm's law formulae V=IR to calculate current.

4 0
3 years ago
What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the
Zepler [3.9K]

Answer:

The magnitude of the magnetic field vector is 1.91T and is directed towards the east.

The steps to the solution can be found in the attachment below.

Explanation:

For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.

Fm = Fg

qvBSin(theta) = mg

Where q = magnitude of charge

v = magnitude of the velocity vector = 4 x10^4 m/s

B = magnitude of the magnetic field vector

theta = the angle between the magnetic field and velocity vectors = 90°

m = mass of the charge = 0.195g

g = acceleration due to gravity =9.8m/s²

On substituting the respective values of all variables in the equation (1) above

B = 1.91T

The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.

Below is the step by step procedure to the solution.

3 0
3 years ago
75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0
3 years ago
HURRYYYWhich tools do meteorologists use to collect data about the weather?
lys-0071 [83]

Answer:

satellites and weather fronts

3 0
3 years ago
Read 2 more answers
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