The return stroke (the current that cause the visible flash) moves upward at a speed of about 320,000,000 ft per second or about 220,000,000 miles per hour (about 1/3 the speed of light).
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Suppose a parachutist lands on a high-voltage wire and grabs the wire as she prepares to be rescued, she will not be electrocuted.
<h3>
What is Voltage?</h3>
This is referred to as the difference in electric potential between two points or the work done which is needed per unit to move a charge between two points in a circuit.
If she grabs the wire as she prepares to be rescued, she will not be electrocuted because the net electrical potential will be 0.
In a situation where the wire breaks as she holds on to it then she will be electrocuted.
Read more about Voltage here brainly.com/question/1176850
#SPJ4
Answer:
x = 3.6 [m]
Explanation:
This problem can be easily solved using a static analysis of forces acting on the ladder, taking into account the respective distances. For easy understanding, a free body diagram should be made.
We perform a sum of force on the X-axis equal to zero, to find that the force exerted by the wall is equal to the friction force on the floor.
Then we perform a summation of forces on the Y axis, to determine that the normal force exerted by the floor is equal to the weight of the ladder.
We know that the friction force is equal to the product of normal force by the coefficient of friction.
In this way, by relating the friction force to the equations deduced above we can find the force exerted by the wall.
Then we make a summation of moments around the base point of the ladder, the equation realized can be seen in the attached image.
In the last analysis we can find the relationship between the horizontal and vertical distance of the ladder, with respect to the wall and the floor.
Then with the complementary analysis of the Pythagorean theorem we can find another additional equation.
The result of the greater distance is 3.6 [m]