Answer:
Explanation:
Electric field talks about a region around a charged particle or object within which a force would be exerted on other charged particles or objects. to find the electric field inside the bulb we will apply the electric filed formula.
Please kindly check attachment for step by step explaination.
(1) A - B
(2) B - C
(3) - A + B - C
(4) 3A - 2C
(5) - 2A + 3B - C
(6) 2A - 3 (B - C)
Answer:
(1) (3,-5,-4)
(2) (-5, 4, 0)
(3) (-6, 4, 3)
(4) (-3, -2, -11)
(5) (-11, 14, 8)
(6) (17, -12, -6)
Explanation:
A⃗ =(1,0,−3)
B⃗ =(−2,5,1)
C⃗ =(3,1,1)
Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.
1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)
2) (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)
3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)
4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)
5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)
6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)
The correct answer to the question is : 29.88 m.
EXPLANATION :
As per the question, the mass of the rock m = 50 Kg.
The rock is rolling off the edges of the cliff.
The final velocity of the rock when it hits the ground v = 24 .2 m/s.
Let the height of the cliff is h.
The potential energy gained by the rock at the top of the cliff = mgh.
Here, g is known as acceleration due to gravity, and g =
When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.
When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.
The kinetic energy of the rock when it touches the ground is given as -
Kinetic energy K.E = .
From above we know that -
Kinetic energy at the bottom of the cliff = potential energy at a height h
⇒
⇒
⇒
⇒
Hence, the height of the cliff is 29.88 m
Here is the complete question
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod
The missing image which is the remaining part of this question is attached in the image below.
Answer:
The horizontal shear stress at point H is
Explanation:
Given that :
The internal shear force V = 80 kN = 80 × 10³ N
The moment of inertia = 64,900,000
The length = 20 mm below the centriod
The horizontal shear stress can be calculated by using the equation:
where;
Q = moment of area above or below the point H
b = thickness of the beam = 10 mm
From the centroid ;
Q =
Q =
Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³
Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³
Q = ( 38500 + 307125 ) mm³
Q = 345625 mm³
The horizontal shear stress at point H is
