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zlopas [31]
4 years ago
6

If a truck weighs 18,000 N and it’s tires are inflated to a pressure of 190 kPa how large is the area of the trucks tires that a

re in contact with the ground
Physics
1 answer:
Mamont248 [21]4 years ago
4 0

0.095m²

Explanation:

Given parameters;

Weight of truck = 18000N

Pressure = 190kPa or 190000Pa

Unknown:

Area of truck tires = ?

Solution:

Pressure is given as the force per unit area of a substance:

            Pressure = \frac{force }{Area}

  Since Area is the unknown, we make it the subject of the expression;

            Area = \frac{force }{pressure}

                     =\frac{18000}{190000}

                     =  0.095m²

Learn more:

Pressure brainly.com/question/7882867

#learnwithBrainly

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An empty capacitor is connected to a 12.0 V battery and charged up. The capacitor is then disconnected from the battery, and a s
Temka [501]

Answer:

7.71 V

Explanation:

When the charge remains constant, the potential difference decrease by the factor of the dielectric constant K

V new = V initial / K = 12 V / 2.8 = 4.29 V

ΔV ( change in potential) = V initial - V new ( after dielectric has been added) = 12 V - 4.29 V =      7.71 V

The potential difference decreases by 7.71 V

3 0
4 years ago
When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde
Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

F=\dfrac{kq_1q_2}{r^2}

r is the separation between charges  

F\propto \dfrac{1}{r^2}

r=\sqrt{\dfrac{1}{F}}

If F'= 2F

r'=\dfrac{1}{\sqrt{2F} }

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r'=\dfrac{1}{\sqrt{2F} }. Hence, this is the required solution.                                                                                    

6 0
3 years ago
A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C
MAVERICK [17]

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

5 0
3 years ago
A student switches the torch on and sees that it gives out a bright light.
garik1379 [7]

Answer:

no more fire duhh

Explanation:

3 0
3 years ago
Read 2 more answers
A sculpture is suspended in equilibrium by two cables, one from a wall and the other
aleksandrvk [35]

Answer:

T_1=6655.295917 \approx 6655.3N

Explanation:

From the question we are told that

Angle of cable 2 \theta=37.0\textdegree

Weight of sculpture W=5000 N

Generally the Tension from cable 2 T_2 is mathematically given by

   T_2sin37\textdegree=5000N

   T_2=5000N/sin37\textdegree

   T_2=8308.2N

Generally the Tension from Cable 1 T_1 is mathematically given by

   T_1=T_2 cos37\textdegree

   T_1=8308.2* cos 37\textdegree

   T_1=6655.295917 \approx 6655.3N

7 0
3 years ago
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