All categories

3 years ago

What is the speed that is measured in speedometer to track speed violation?

Physics

1 answer:

BartSMP [9]3 years ago

4 0

Answer:

The officer's unit detects this 135-mile-per-hour speed and should subtract the patrol car's 70-mile -per-hour ground speed to get your true speed of 65 miles per hour. Instead, the officer's ground-speed beam fixes on the truck ahead and measures a false 50-mile-per-hour ground speed.

Explanation:

A speedometer or speed meter is a gauge that measures and displays the instantaneous speed of a vehicle. Now universally fitted to motor vehicles, they started to be available as options in the early 20th century, and as standard equipment from about 1910 onwards.

You might be interested in

What is zero on the kelvin scale

elena55 [62]

The point at which all motion stops.

8 0

3 years ago

Read 2 more answers

You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of

Sunny_sXe [5.5K]

Answer:

$P = 1.090\,N$

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

$\Sigma F = P - W + F_{D}$

$\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0$

Now, the exerted force is:

$P = \rho \cdot V \cdot g - c\cdot v$

The volume of a sphere is:

$V = \frac{4\cdot \pi}{3}\cdot R^{3}$

$V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}$

$V = 1.149\times 10^{-5}\,m^{3}$

Lastly, the force is calculated:

$P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )$

$P = 1.090\,N$

5 0

3 years ago

A little girl is swinging on her tire swing. At the lowest point of her swing, the tension in the rope is a) Equal to the gravit

Elodia [21]

A because the girl in that instant is not moving up or down so
( up forces)=(down forces )
The up forces is the tension of the rope and down forces us mg -the gravitational force on the girl by the earth

6 0

3 years ago

Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re

ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of $62 lb / ft ^ 3$ (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

$W = \gamma A * \int_0^15 dy$

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

$W = (62)(14*7)\int^{15}_0 (15-y)dy$