The magnitude of the electric field at the proton's location is 10,437.5 N/C.
<h3>What the magnitude of the
electric field?</h3>
The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.
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Answer:
94
Explanation:
f = 2.57 x 10^13 Hz
E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J
Energy of each photon = h f
Where, h is Plank's constant
Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J
Number of photons = Total energy / energy of one photon
N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94
The time taken for the tiny saliva to travel is 0.55 second.
The horizontal distance traveled at speed of 4 m/s is 2.2 m.
The horizontal distance traveled at speed of 20 m/s is 11 m.
<h3>
Time of motion of the tiny saliva</h3>
The time taken for the tiny saliva to travel is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- g is the acceleration due to gravity
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
Substitute the given parameters and solve for time of motion;
t = √(2 x 1.5 / 10)
t = 0.55 second
<h3>Horizontal distance traveled at speed of 4 m/s</h3>
X = Vx(t)
X = (4 m/s)(0.55)
X = 2.2 m
<h3>Horizontal distance traveled at speed of 20 m/s</h3>
X = (20)(0.55)
X = 11 m
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Answer:
The correct answer to the question is
Both A and B are true
Explanation:
The particles of a gas are free to move to occupy the entire volume in which they are placed due to the smallerinter molecular forces holding them together hence due to the face that pressure is a measure of the Force per unit area that is Pressure P = ( Force F)/ (Area A) then the force per unit area, exerted on the all of the container by the gaseous particles which are colliding with each other and with the walss of the container is fairly constant through out the surface oof the container
In the case of the liquid which are held on together by more stronger forces, the force per nit area exerted by the liquid particle is transmitted from one particle to the next until it reaches the container's surface. Then remembering that the force of gravity on the liquid is acting in one direction (that is downwards) the sum of the fprce due to the weight incrreases as we progress deaper into the liquid hence the pressure increases per unit depth
Answer:
50 N
4.2 N
Explanation:
i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.
ii) When the boom is resting on the end support, the normal force is:
∑τ = Iα
-W (0.50) + F (3.0) − N (6.0) = 0
-0.50 W + 3.0 F = 6.0 N
N = (-0.50 W + 3.0 F) / 6.0
N = (-0.50 × 2350 + 3.0 × 400) / 6.0
N ≈ 4.2
