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Whitepunk [10]
3 years ago
8

The force exerted by electrically changed objects

Physics
1 answer:
MAVERICK [17]3 years ago
8 0

Answer:

Electrostatic Force...or a non-contact force

Explanation:

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3 years ago
Which of the following statements are true about the international system of measurement?
anygoal [31]
The International System Units or the SI units is  scientific method of expressing the magnitudes or quantities of important natural phenomena. There are seven base units in the system, from which other units are derived. This system was formerly called the meter-kilogram-second (MKS) system.
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3 years ago
Three laws that indicate a chemical change
alexandr1967 [171]

Temperature, The highness, and the time.

Hope this helps!

C=

4 0
3 years ago
An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a
prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611

Converting to SI units

10\ ft/s=10\times \dfrac{1}{3.281}

32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}

2\ ft=2\times \dfrac{1}{3.281}

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

7 0
3 years ago
A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a
MrRissso [65]

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

6 0
3 years ago
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