A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw b
2 answers:
Answer:
A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
Explanation:
As we know that electrostatic force on a charge in external electric field is given as
so here the force and electric field will be in same direction if charge is positive and if the charge is negative then force will be in opposite direction to the electric field.
So we know that since the electron is negatively charged so the force on electron must be opposite to the electric field present in that region.
So correct answer will be
A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
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Answer:
a) a = g / 3
b) x (3.0) = 14.7 m
c) m (3.0) = 29.4 g
Explanation:
Given:-
- The following differential equation for (x) the distance a rain drop has fallen has the form:
- Where, v = Speed of the raindrop
- Proposed solution to given ODE:
v = a*t
Where, a = acceleration of raindrop
Find:-
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
Solution:-
- We know that acceleration (a) is the first derivative of velocity (v):
a = dv / dt ... Eq 1
- Similarly, we know that velocity (v) is the first derivative of displacement (x):
v = dx / dt , v = a*t ... proposed solution (Eq 2)
v .dt = dx = a*t . dt
- integrate both sides:
∫a*t . dt = ∫dt
x = 0.5*a*t^2 ... Eq 3
- Substitute Eq1 , 2 , 3 into the given ODE:
0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2
= 1.5 a^2 t^2
a = g / 3
- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:
x (t) = 0.5*a*t^2
x (t = 3.0) = 0.5*9.81*3^2 / 3
x (3.0) = 14.7 m
- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:
m (t) = k*x (t)
m (3.0) = 2.00*x(3.0)
m (3.0) = 2.00*14.7
m (3.0) = 29.4 g
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
