Answer:
6.44 × 10^10 N/C
Explanation:
Electric field due to the ring on its axis is given by
E = K q r / (r^2 + x^2)^3/2
Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.
r = 0.25 m, x = 0.5 m, q = 5 C
K = 9 × 10^9 Nm^2/C^2
E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2
E = 6.44 × 10^10 N/C
The value of cos θ in the given figure is 0.98.
<h3>
What is cosine of an angle?</h3>
The cosine of an angle is defined as the sine of the complementary angle.
The complementary angle equals the given angle subtracted from a right angle, 90.
cos θ = sin(90 - θ)
For example, if the angle is 30°, then its complement is 60°
cos 30 = sin(90 - 30)
cos 30 = sin 60
0.866 = 0.866
<h3>Cosine of an angle with respect to sides of a right triangle</h3>
cos θ = adjacent side / hypotenuse side
adjacent side of the given right triangle is calculated as follows;
adj² = 10² - 2²
adj² = 100 - 4
adj² = 96
adj = √96
adj = 9.8
cos θ = 9.8/10
cos θ = 0.98
Thus, the value of cos θ in the given figure is 0.98.
Learn more about cosine of angles here: brainly.com/question/23720007
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Answer:
The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
Putting values in above equation, we get:
The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
The solution for this problem is:
If they feel 50% of their weight that means that the
centripetal force is also 50% of their weight 1g - 0.5g = 0.5g
Then 0.5* 9.8m/s² * 18m = 88.2 would be v²
Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.
Answer:
865.08 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 243 m/s
Height (h) of the cliff = 62 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:
Height (h) of the cliff = 62 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
62 = ½ × 9.8 × t²
62 = 4.9 × t²
Divide both side by 4.9
t² = 62/4.9
Take the square root of both side.
t = √(62/4.9)
t = 3.56 s
Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:
Initial velocity (u) = 243 m/s
Time (t) = 3.56 s
Horizontal distance (s) =?
s = ut
s = 243 × 3.56 s
s = 865.08 m
Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.
