What is the difference between 3.15 m and 2.006 m with the correct number of significant figures
2 answers:
Answer:
S = 1.144
Explanation:
As we know that when we add or subtract two or more numbers then the final result must have same number of precision as that of numbers given in the problem
here we have
S = 3.15 - 2.006
so we have to give our result with 3 digits after decimal
so we will have
since it must have more precise digits so correct answer after subtraction must be
S = 1.144
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Answer:
665 ft
Explanation:
Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.
The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is
Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees
Since the 2 sides length above make up the 200 foot monument, their total length is
0.231d + 0.07d = 200
0.301 d = 200
d = 200 / 0.301 = 665 ft
Answer:
Figure a. E_net = 99.518 N/C
Figure b. E_net = 177.151 N / C
Explanation:
Given:
- Attachment for figures missing in the question.
- The dimensions for rectangle are = 7.79 x 3.99 cm
- All four charges have equal magnitude Q = 10.6*10^-12 C
Find:
Find the magnitude of the electric field at the center of the rectangle in Figures a and b.
Solution:
- The Electric field generated by an charged particle Q at a distance r is given by:
E = k*Q / r^2
- Where, k is the coulomb's constant = 8.99 * 10^9
Part a)
- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:
E_1 + E_3 = 0
- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:
E_net = E_2 + E_4
E_2 = E_4
E_net = 2*E = 2*k*Q / r^2
- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:
r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )
r = sqrt ( 1.9151*10^-3 ) = 0.043762 m
- Plug the values in the E_net expression developed above:
E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3
E_net = 99.518 N/C
Part b)
- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).
- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.
Hence,
E_net = 2*E_part(a)*cos(Q)
- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:
Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.
- Now, compute the net electric field E_net:
E_net = 2*(99.518)*cos(27.12)
E_net = 177.151 N / C
Answer:
2560m or 2.56km (rounded to 3 significant figures)
Explanation:
First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):
The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;
First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;
Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);
What we have is the horizontal velocity but we don't have the time taken;
One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;
What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:
800 = (0)t + ¹/₂·(9.8)t²
800 = 4.9t²
t² = 163.26...
t = 12.77...
We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:
Answer:
Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)
1) Todas las fuerzas están en la misma dirección.
Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:
F = 48N + 60N + 30N = 138N
Y por la segunda ley de Newton sabemos que:
F = m*a
fuerza igual a masa por aceleración.
Entonces la aceleración está dada por:
a = F/m = 138N/12kg = 11.5 m/s^2
2) Segundo caso, suponemos que F1 es opuesta a F2 y F3
En este caso, la fuerza neta será:
F = F2 + F3 - F1 = 60N + 30N - 48N = 42N
En este caso, la aceleración será:
a = 42N/12kg = 3.5 m/s^2
