Answer:
1.758820×10^11(-2.5i-0.8j) m/s^2
Explanation:
From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...
We can solve this problem using the formula below;
Ma= q[E+V × B] ---------------(1).
Note: q is negative, m= mass of electron.
Making acceleration,a the subject of the formula and substituting the parameters into equation (1);
a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)
a= -e/m × (2.5i+5.2j-4.4j)
a= e/m × (-2.5i - 0.8j)
e/m= 1.758820×10^11 c/kg
Therefore, slotting in the value of charge to mass(e/m) ratio;
a= 1.7588×10^11×(-2.5i-0.8j) m/s^2
<h2>Answers:</h2><h2 /><h2>a) Arrow B</h2><h2>b) Arrow E</h2>
Explanation:
Refraction is a phenomenon in which a wave (the light in this case) bends or changes its direction <u>when passing through a medium with a refractive index different from the other medium.</u> Where the Refractive index is a number that describes how fast light propagates through a medium or material.
According to this, if we observe the rays A an D passing throgh the biconcave lens, we will have two mediums:
1) The air
2)The material of the biconcave lens
This two mediums have different refractive indexes, hence the rays will change the direction.
-For the incident ray A, the corresponding refractive ray is B, because is the ray that bends after passing throgh the lens
-For the incident ray D, the refracted ray is E following the same principle.
A. True
Hope this helps :)
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
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