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jok3333 [9.3K]
3 years ago
7

PLEASE HURRY WILL GIVE BRAINLIEST four people went out for a bike ride. which had the greatest average speed?

Physics
1 answer:
klasskru [66]3 years ago
4 0
The answer would be c. hope this helps☺





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3 years ago
You will measure and record the time it takes the cart to travel certain distances, and then complete some calculations. In the
natka813 [3]

Answer:

Take the measurement of the distance (d) with a meter rule (in meters) and also measure the time (t) of the travel in seconds with a stopwatch.

question: What is the speed of the cart?

Explanation:

The speed of an object in motion is the distance covered by the object with respect to time, that is, the ratio of distance covered to the time taken to reach that distance.

Speed = distance / time

           = d (in meters m) / t (in seconds s) = m/s

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2 years ago
A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determi
KatRina [158]

Answer:

Explanation is given

Mark Me as Branliest

3 0
3 years ago
A bike rider pedals with constant acceleration to reach a velocity of 7.8 m/s over a time of 4.2 s. during the period of acceler
Artyom0805 [142]

To calculate the initial velocity of the bike, we use the following equation

d=\frac{1}{2} (u+v)t.

or

u=\frac{2d}{t} -v

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.

Given, u =7.8 m/s,d= 19 m and t=4.2 s.

Substituting these values in above equation, we get

u = \frac{2 \times 19}{4.2 \ s} -7.8 m/s = 9.05 \ m/s - 7.8 \ m/s \\\\ u= 1.2 m/s.

Thus, the initial velocity of the bike is 1.2 m/s.

3 0
3 years ago
Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres
inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )

Here,  v_{1} is the velocity of water through wide ends of cylindrical pipe and v_{2} is the velocity of water through narrow ends of cylindrical pipe.

Given, v_{1} =1.4 m/s

Now from equation continuity,

v_{1} A_{1} = v_{2} A_{2}.

Here, A_{1} and A_{2} are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}.

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}

v_{2} = 4 \times 1.4 = 5.6 m/s

Substituting these values  with the density of water is 1000 \ kg/m^3 in pressure difference formula we get.

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa

3 0
2 years ago
Read 2 more answers
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