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2 years ago

Which physical activity would a global positioning system resource help you with?

Physics

1 answer:

PolarNik [594]2 years ago

4 0

The physical activity that a global positioning system resource can help me with is : ( A ) Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

GPS helps with the accurate measurement of physical activities and factors such as location, time, elevation and so on. When hiking, the time, distance covered and location of the hiker can be accurately measured with the use of the GPS.

Hence we can conclude that The physical activity that a global positioning system resource can help me with is Hiking.

Learn more about GPS : brainly.com/question/9795929

#SPJ1

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If the Fox for an object on an incline is 350N,

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3 years ago

Which oftthe following is the least reliable source of background information for a scientific project? General internet site, g

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General internet site

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3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago

As the wavelength increases, the frequency (2 points) Select one:

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A.
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4 years ago

Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state

topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

The relation of the process P = constant

Pressure of point (1) P1 = 10 bar = P2

Volume of point (1) V1 = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant

Process 2 ---> 3

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

Process 3 ---> 1 

The relation of the process PV = constant

Required

Sketch the processes on the PV coordinates

The work for each process in kJ

Solution

The work is defined by

W= $\int\limits^a_b {x} \, dx$

a=V2

b=V1

x=P

dx=dV

Process 1 ---> 2

P3 = P4 = 5 bar

W= $\int\limits^a_b {x} \, dx$

a=V3

b=V2

x=4

dx=dV

putting the value of a, b, x, dx in above integral

W=400 kJ

Process 2 ---> 3 

V = constant Then there is no change in the volume,hence W = 0 kJ

Process 3 ---> 1

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1

W= $\int\limits^a_b {x} \, dx$

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

= -160.944 KJ

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3 years ago